MH1810 Math

Chapter 5c: Differentiation

Imron Rosyadi

5.8 Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental theorem that connects the average rate of change of a function over an interval to its instantaneous rate of change at some point within that interval.

Theorem 5.8.1 (Rolle’s Theorem):

Let \(f\) be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). If \(f(a) = f(b)\), then there is at least one point \(c\) in \((a, b)\) such that \(f'(c) = 0\).

(Geometrically: If a function starts and ends at the same height, and is smooth, its tangent must be horizontal somewhere in between.)

Theorem 5.8.2 (The Mean Value Theorem):

Let \(f\) be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). Then there is (at least one point) \(c\) in \((a, b)\) such that \[ \frac {f (b) - f (a)}{b - a} = f ^ {\prime} (c). \]

(Geometrically: There is a point \(c\) where the tangent line is parallel to the secant line connecting \((a, f(a))\) and \((b, f(b))\).)

The Mean Value Theorem is one of the most significant results in calculus. It’s an “existence” theorem, meaning it guarantees that something exists, even if it doesn’t tell you how to find it.

Rolle’s Theorem is a special case of the MVT. It says that if a function is continuous and differentiable, and has the same value at two endpoints, then there must be at least one point in between where the derivative is zero (a horizontal tangent). Think of it as reaching a peak or a valley if you start and end at the same elevation.

The Mean Value Theorem generalizes this. It states that if a function satisfies the conditions of continuity and differentiability on an interval, then there is at least one point \(c\) within that interval where the instantaneous rate of change (\(f'(c)\)) is equal to the average rate of change over the entire interval (\(\frac{f(b)-f(a)}{b-a}\)). Geometrically, this means there’s a tangent line parallel to the secant line connecting the endpoints. This theorem is crucial for many proofs in calculus, including some of the upcoming theorems about function behavior.

Applications of Mean Value Theorem

Example 5.8.3: Bounding Function Values

Suppose \(f(0) = -3\) and \(f'(x) \leq 5\) for all \(x\). How large can \(f(2)\) be?

Solution:

Since \(f'(x)\) exists for all \(x\), \(f\) is differentiable and therefore continuous everywhere.

Apply the Mean Value Theorem to \(f\) on the interval \([0, 2]\).

There exists some \(c \in (0, 2)\) such that: \[ \frac {f (2) - f (0)}{2 - 0} = f ^ {\prime} (c). \] We are given \(f'(x) \leq 5\), so \(f'(c) \leq 5\). \[ \frac {f (2) - (-3)}{2} \leq 5. \] \[ f (2) + 3 \leq 10. \] \[ f (2) \leq 7. \] The largest value that \(f(2)\) can have is 7.

Applications of Mean Value Theorem

Example 5.8.4: Estimating with MVT

Use the Mean Value Theorem to estimate \(\sqrt[3]{65}\).

Solution:

Let \(f(x) = \sqrt[3]{x} = x^{1/3}\). We want to estimate \(f(65)\).

Choose the interval \([64, 65]\) because \(64\) is a perfect cube and close to \(65\).

\(f(x)\) is continuous on \([64, 65]\) and differentiable on \((64, 65)\).

\(f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}\).

By MVT, there is an \(x_0 \in (64, 65)\) such that: \[ \frac {f (6 5) - f (6 4)}{6 5 - 6 4} = f ^ {\prime} (x _ {0}). \] \[ \sqrt [ 3 ]{6 5} - \sqrt [ 3 ]{6 4} = \frac {1}{3 x _ {0} ^ {2 / 3}}. \] \[ \sqrt [ 3 ]{6 5} - 4 = \frac {1}{3 x _ {0} ^ {2 / 3}}. \] \[ \sqrt [ 3 ]{6 5} = 4 + \frac {1}{3 x _ {0} ^ {2 / 3}}, \text {w h e r e} x _ {0} \in (6 4, 6 5). \]

Now, estimate \(\frac{1}{3x_0^{2/3}}\). Since \(64 < x_0 < 65\):

\(x_0^{2/3} > 64^{2/3} = (\sqrt[3]{64})^2 = 4^2 = 16\).

So, \(3x_0^{2/3} > 3(16) = 48\).

Therefore, \(\frac{1}{3x_0^{2/3}} < \frac{1}{48}\).

Thus, \(\sqrt[3]{65} < 4 + \frac{1}{48}\).

We know \(f(64) = 4\), so \(4 < \sqrt[3]{65}\).

Combining, \(4 < \sqrt[3]{65} < 4 + \frac{1}{48} \approx 4.02083\).

(Actual value \(\sqrt[3]{65} \approx 4.020726...\))

The Mean Value Theorem has both theoretical and practical applications.

In the first example, MVT helps us bound the value of a function. Given \(f(0)=-3\) and a bound on its derivative (\(f'(x) \le 5\)), we can use the MVT to determine the maximum possible value for \(f(2)\). This is a common type of problem in bounding errors or changes in systems.

The second example shows how MVT can be used for estimation, similar to linearization but providing an interval for the approximation rather than a single value. To estimate \(\sqrt[3]{65}\), we choose a nearby point where the cube root is easily calculated (\(x=64\)). Applying MVT gives us an expression for \(\sqrt[3]{65}\) involving \(f'(x_0)\) for some \(x_0\) between 64 and 65. By bounding \(f'(x_0)\) (since \(x_0\) is between 64 and 65, we know \(x_0^{2/3}\) must be greater than \(64^{2/3}\)), we can create an interval within which \(\sqrt[3]{65}\) must lie. This provides a rigorous way to estimate values and bound their uncertainty.

5.8.1 L’Hospital’s Rule

L’Hospital’s Rule is a powerful tool for evaluating indeterminate forms of limits.

Indeterminate Forms:

Limits of fractions where both numerator and denominator tend to zero (type \(\frac{0}{0}\)) or both tend to \(\pm \infty\) (type \(\frac{\pm\infty}{\pm\infty}\)).

• \(\lim_{x\to 1}\frac{\ln x}{x - 1}\) (\(\frac{0}{0}\))
• \(\lim_{x\to 1^{+}}\frac{x^3 - 1}{\sqrt{x - 1}}\) (\(\frac{0}{0}\))
• \(\lim_{x\to \infty}\frac{e^x}{x^2}\) (\(\frac{\infty}{\infty}\))

Theorem 5.8.6 (L’Hospital’s Rule):

Suppose \(f\) and \(g\) are differentiable, and \(g(x)\), \(g'(x)\) are non-zero near \(a\) (except possibly at \(a\)).

If \(\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0\) (type \(\frac{0}{0}\)), OR

If \(\lim_{x \to a} f(x) = \pm \infty\) and \(\lim_{x \to a} g(x) = \pm \infty\) (type \(\frac{\pm\infty}{\pm\infty}\)),

Then:

\[ \lim _ {x \to a} \frac {f (x)}{g (x)} = \lim _ {x \to a} \frac {f ^ {\prime} (x)}{g ^ {\prime} (x)} \]

(if the latter limit exists or diverges to \(\infty\) or \(-\infty\)).

(The theorem also holds for one-sided limits and limits at infinity.)

Warning

Conditions MUST be satisfied! Do NOT apply L’Hospital’s Rule if the limit is not an indeterminate form.

Example: \(\lim_{x \to 1} \frac{x+1}{x} = \frac{1+1}{1} = 2\).

If you wrongly apply L’Hospital’s: \(\lim_{x \to 1} \frac{1}{1} = 1\), which is incorrect.

L’Hospital’s Rule is one of the most powerful tools for evaluating limits that are in “indeterminate forms.” These are situations where direct substitution gives us expressions like \(0/0\) or \(\infty/\infty\), which don’t immediately tell us the limit’s value.

The rule states that if your limit is an indeterminate form, you can take the derivative of the numerator and the derivative of the denominator separately, and then evaluate the limit of that new fraction. If that new limit exists, it’s the same as the original limit.

The proof provided for the \(\frac{0}{0}\) case for continuous \(f'\) and \(g'\) functions uses the definition of the derivative, highlighting its connection.

It’s absolutely critical to check the conditions before applying L’Hospital’s Rule. If the limit is not an indeterminate form, applying the rule will give you the wrong answer. The warning example clearly illustrates this common mistake. Always verify the indeterminate form first!

Examples of L’Hospital’s Rule

Example 5.8.7: Find the limit \(\lim _ {x \to 1} {\frac {\ln x}{x - 1}}\).

Solution:

As \(x \to 1\), \(\ln x \to \ln 1 = 0\) and \(x - 1 \to 1 - 1 = 0\). This is an indeterminate form of type \(\frac{0}{0}\).

Apply L’Hospital’s Rule: \[ \lim_{x\to 1}\frac{\ln x}{x - 1}\underbrace{=}_{L^{\prime}H}\lim_{x\to 1}\frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x - 1)} = \lim_{x\to 1}\frac{\frac{1}{x}}{1} = \frac{1/1}{1} = 1. \]

Example 5.8.8: Evaluate the limit \(\lim_{x\to \infty}\frac{e^x}{x^2}\).

Solution:

As \(x \to \infty\), \(e^x \to \infty\) and \(x^2 \to \infty\). This is an indeterminate form of type \(\frac{\infty}{\infty}\).

Apply L’Hospital’s Rule (potentially repeatedly): \[ \underbrace{\lim_{x\to\infty}\frac{e^{x}}{x^{2}}}_{\substack{\infty / \infty}}\underbrace{=}_{L^{\prime}H}\underbrace{\lim_{x\to\infty}\frac{e^{x}}{2x}}_{\substack{\infty / \infty}}\underbrace{=}_{L^{\prime}H}\lim_{x\to \infty}\frac{e^{x}}{2} = \infty . \]

(Since \(e^x\) grows much faster than any polynomial.)

Question: Would you apply L’Hospital’s Rule to the following?

1. \(\lim _ {x \to \infty} \frac {\sqrt {x ^ {2} + 2}}{\sqrt {x ^ {2} + 5}}\)? (Answer: Yes, it’s \(\frac{\infty}{\infty}\), but simpler with algebraic manipulation)

2. \(\lim _ {x \rightarrow \infty} \frac {x ^ {1 7 9} + x ^ {1 7 8} + \cdots + x + 1}{3 x ^ {1 7 9} - 2 x ^ {1 7 8} + \cdots + 3 x - 2}\)?

   (Answer: Yes, it’s \(\frac{\infty}{\infty}\), but simpler by dividing by highest power of x)

These examples demonstrate the application of L’Hospital’s Rule to both \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\) forms.

For \(\lim_{x \to 1} \frac{\ln x}{x-1}\), direct substitution gives \(0/0\). Applying L’Hospital’s rule by taking derivatives of the numerator and denominator simplifies the expression significantly, leading to the limit value of 1.

For \(\lim_{x \to \infty} \frac{e^x}{x^2}\), direct substitution gives \(\infty/\infty\). We apply L’Hospital’s rule, and we find that it’s still \(\infty/\infty\). This means we can apply the rule again. After two applications, the denominator becomes a constant, and the limit clearly goes to infinity. This illustrates that you can apply L’Hospital’s rule multiple times as long as the indeterminate form persists.

The questions at the end are important for critical thinking. While L’Hospital’s rule could be applied to these limits (they are \(\infty/\infty\) forms), other methods (like dividing by the highest power of \(x\) for rational functions or algebraic simplification) are often much faster and less error-prone. L’Hospital’s rule is a powerful tool, but it’s not always the most efficient one.

5.8.2 Other Indeterminate Forms

L’Hospital’s rule directly applies only to \(\frac{0}{0}\) and \(\frac{\pm\infty}{\pm\infty}\). Other indeterminate forms often require algebraic manipulation to transform them into one of these forms.

Common Indeterminate Forms:

• \(0 \cdot \infty\)
• \(\infty - \infty\)
• \(1^\infty\)
• \(0^0\)
• \(\infty^0\)

5.8.2 Other Indeterminate Forms

Example 5.8.9: Evaluate the limit \(\lim _ {x \to 0 ^ {+}} x \ln x\).

This is an indeterminate form of type \(0 \cdot (-\infty)\).

Solution:

Rewrite \(x \ln x\) as a quotient:

\[ x \ln x = \frac {\ln x}{1/x}. \]

Now the limit is \(\lim _ {x \to 0 ^ {+}} \frac {\ln x}{1/x}\), which is of type \(\frac{-\infty}{\infty}\).

Apply L’Hospital’s Rule: \[ \lim _ {x \to 0 ^ {+}} \frac {\ln x}{1 / x} \underbrace {=}_{L^{\prime}H} \lim _ {x \to 0 ^ {+}} \frac {1 / x}{- 1 / x ^ {2}} = \lim _ {x \to 0 ^ {+}} (- x) = 0. \]

(Alternatively, we could write \(x \ln x = \frac{x}{1/\ln x}\), but differentiating \(1/\ln x\) is more complex.)

Example 5.8.10: Evaluate \(\lim _ {x \to 0 ^ {+}} \left(x ^ {x}\right)\).

This is an indeterminate form of type \(0^0\).

Solution:

Use the property \(A^B = e^{\ln(A^B)} = e^{B \ln A}\).

So, \(x^x = e^{\ln(x^x)} = e^{x \ln x}\).

\[ \lim _ {x \to 0 ^ {+}} \left(x ^ {x}\right) = \lim _ {x \to 0 ^ {+}} \exp (x \ln x). \]

Since \(\exp(u)\) is a continuous function, we can move the limit inside the exponent:

\[ \exp \left( \lim_{x\to 0^{+}}(x\ln x) \right). \]

From the preceding example, we know \(\lim _ {x \to 0 ^ {+}} x \ln (x) = 0\).

Therefore, \(\lim _ {x \rightarrow 0 ^ {+}} (x ^ {x}) = \exp (0) = 1\).

L’Hospital’s Rule is directly applicable only to quotients. However, other indeterminate forms can often be cleverly rewritten as quotients.

For \(0 \cdot \infty\), like \(\lim_{x \to 0^+} x \ln x\), the trick is to move one of the factors to the denominator as its reciprocal. For example, \(x \ln x = \frac{\ln x}{1/x}\). This transforms it into an \(\frac{\infty}{\infty}\) form, allowing us to apply L’Hospital’s Rule. We then proceed with the differentiation and evaluation of the limit.

For forms like \(0^0\), \(1^\infty\), or \(\infty^0\), the strategy involves using logarithms. By taking the natural logarithm, we convert the exponential form into a product, specifically \(B \ln A\). This usually transforms the problem into a \(0 \cdot \infty\) form, which we then convert into a quotient, as in the previous case. For \(\lim_{x \to 0^+} x^x\), after taking the logarithm, we get \(\lim_{x \to 0^+} x \ln x\), which we just solved. Since the exponential function is continuous, we can take the limit of the exponent first and then exponentiate the result. This leads us to the surprising result that \(0^0 = 1\) in this context. These examples show the power of algebraic manipulation combined with L’Hospital’s Rule for a broader range of limit problems.

5.9 Maximum and Minimum Problems

The first derivative can tell us if a function is increasing or decreasing, and help locate extrema.

Theorem 5.9.1: First Derivative Test for Monotonicity

1. If \(f'(x) > 0\) on \((a, b)\), then \(f\) is increasing on \((a, b)\).

   (i.e., for \(x_1 < x_2 \in (a, b)\), then \(f(x_1) < f(x_2)\)).

2. If \(f'(x) < 0\) on \((a, b)\), then \(f\) is decreasing on \((a, b)\).

   (i.e., for \(x_1 < x_2 \in (a, b)\), then \(f(x_1) > f(x_2)\)).

(Proof uses Mean Value Theorem)

Corollary 5.9.2:

Suppose \(f\) is continuous on \([a, b]\).

1. If \(f'(x) > 0\) on \((a, b)\), then \(f\) is increasing on \([a, b]\).
2. If \(f'(x) < 0\) on \((a, b)\), then \(f\) is decreasing on \([a, b]\).

5.9 Maximum and Minimum Problems

Example 5.9.3: Increasing Interval

Find interval(s) where \(f(x) = 2 + 3x - x^3\) is increasing.

Solution:

1. Find \(f'(x)\): \(f'(x) = 3 - 3x^2 = 3(1 - x^2) = 3(1 - x)(1 + x)\).

2. Find where \(f'(x) > 0\):

   We need \(3(1 - x)(1 + x) > 0\).

   This occurs when \((1 - x)\) and \((1 + x)\) have the same sign.

   • If \(1 - x > 0\) and \(1 + x > 0 \Rightarrow x < 1\) and \(x > -1 \Rightarrow -1 < x < 1\).
   • If \(1 - x < 0\) and \(1 + x < 0 \Rightarrow x > 1\) and \(x < -1\) (no solution).

Therefore, \(f(x)\) is increasing on the interval \((-1, 1)\).

Since \(f\) is continuous, by the corollary, \(f\) is increasing on \([-1, 1]\).

The first derivative is a powerful tool for analyzing the behavior of functions. It directly tells us about monotonicity: whether a function is increasing or decreasing.

Theorem 5.9.1 states that if the derivative is positive, the function is increasing; if the derivative is negative, the function is decreasing. This is quite intuitive: a positive slope means the function is going “uphill,” and a negative slope means it’s going “downhill.” The proof for this relies on the Mean Value Theorem.

The corollary extends this to closed intervals, provided the function is continuous. This is important for practical applications.

In Example 5.9.3, we find the derivative of the polynomial \(f(x) = 2 + 3x - x^3\). By factoring the derivative, we can easily find the intervals where \(f'(x) > 0\). We use a sign chart or test points to determine where the derivative is positive. This analysis tells us that the function is increasing between \(x=-1\) and \(x=1\). Such analysis is crucial for understanding a function’s graph and for optimization problems.

Using \(f'\) to Identify One-to-One Functions

If a function is strictly increasing or strictly decreasing on an interval, then it is one-to-one on that interval.

Example 5.9.4: Show that \(f(x) = \sin x\) with domain \([-\pi/2, \pi/2]\) is one-to-one.

Solution:

1. Find \(f'(x)\): \(f'(x) = \cos x\).

2. Analyze \(f'(x)\) on the given interval: For \(x \in (-\pi/2, \pi/2)\), \(\cos x > 0\).

3. Conclusion: Since \(f(x) = \sin x\) is continuous on \([-\pi/2, \pi/2]\) and \(f'(x) > 0\) on \((-\pi/2, \pi/2)\), by Corollary 5.9.2, \(f\) is strictly increasing on \([-\pi/2, \pi/2]\).

   Therefore, \(f(x) = \sin x\) is one-to-one on this interval.

   (This is why its inverse, \(\sin^{-1}x\), exists with a well-defined domain and range.)

Using \(f'\) to Solve Optimization Problems (Fermat’s Theorem)

Theorem 5.9.5 (Fermat’s Theorem):

Suppose \(f\) has a local maximum or minimum at \(c\). If \(f'(c)\) exists, then \(f'(c) = 0\).

Important

This means that if a local extremum occurs at a point where the derivative exists, that point must be a stationary point (\(f'(c)=0\)). This is why we seek critical points when finding extrema!

The first derivative test for monotonicity is not just for graphing; it also helps us understand injectivity (one-to-one property) of functions. If a function is always increasing or always decreasing over an interval, it will never take on the same \(y\)-value twice, hence it is one-to-one.

Example 5.9.4 uses this idea for \(f(x) = \sin x\) on the restricted domain \([-\pi/2, \pi/2]\). Its derivative, \(\cos x\), is positive on \((-\pi/2, \pi/2)\), which means \(\sin x\) is strictly increasing there. This explains why \(\arcsin x\) can be defined as a proper inverse function on this interval.

Fermat’s Theorem is a cornerstone of optimization. It tells us that if a function has a local maximum or minimum at a point where it’s differentiable, then the derivative must be zero at that point. Geometrically, this means the tangent line is horizontal. This theorem is why “finding where \(f'(x)=0\)” is the first step in many optimization problems. It identifies the “stationary points” where peaks and valleys (local extrema) can occur. It’s a necessary condition for local extrema, but not sufficient (i.e., \(f'(c)=0\) doesn’t guarantee an extremum, like at an inflection point).

Example: Optimizing Surface Area

Example 5.9.6: Construct a cylindrical metal can with a given volume \(V\) that minimizes the surface area.

Solution:

Let \(r\) be the radius and \(h\) the height.

• Volume: \(V = \pi r^2 h\).
• Surface Area: \(A = 2\pi r^2 + 2\pi r h\).

Our goal is to minimize \(A\). We need to express \(A\) as a function of a single variable.

From the volume equation, \(h = \frac{V}{\pi r^2}\). Substitute this into the area equation: \[ A(r) = 2\pi r^2 + 2\pi r \left(\frac{V}{\pi r^2}\right) = 2\pi r^2 + \frac{2V}{r}. \] The domain for \(r\) is \((0, \infty)\) (radius must be positive). \(A(r)\) is continuous on this domain.

Example: Optimizing Surface Area (cont.)

1. Find \(A'(r)\):

   \(A'(r) = \frac{d}{dr}(2\pi r^2 + 2Vr^{-1}) = 4\pi r - 2Vr^{-2} = 4\pi r - \frac{2V}{r^2}\).

   Rewrite to find critical points:

   \(A'(r) = \frac{4\pi r^3 - 2V}{r^2} = \frac{2(2\pi r^3 - V)}{r^2}\).

2. Find Critical Points:

   • Singular points: \(A'(r)\) is undefined at \(r=0\), but \(r=0\) is not in the domain \((0, \infty)\).

   • Stationary points: \(A'(r) = 0\) when the numerator is zero.

     \(2\pi r^3 - V = 0 \Rightarrow r^3 = \frac{V}{2\pi} \Rightarrow r = \left(\frac{V}{2\pi}\right)^{1/3}\).

     This is our only critical point in the domain \((0, \infty)\).

3. Analyze using First Derivative Test (Sign Chart):

   • If \(0 < r < \left(\frac{V}{2\pi}\right)^{1/3}\), then \(r^3 < \frac{V}{2\pi}\), so \(2\pi r^3 - V < 0\).

     Thus, \(A'(r) = \frac{\text{negative}}{\text{positive}} < 0\). So \(A(r)\) is decreasing.

   • If \(r > \left(\frac{V}{2\pi}\right)^{1/3}\), then \(r^3 > \frac{V}{2\pi}\), so \(2\pi r^3 - V > 0\).

     Thus, \(A'(r) = \frac{\text{positive}}{\text{positive}} > 0\). So \(A(r)\) is increasing.

   Since \(A(r)\) changes from decreasing to increasing at \(r = \left(\frac{V}{2\pi}\right)^{1/3}\), this critical point corresponds to a global minimum.

4. Find the optimal height:

   \(h = \frac{V}{\pi r^2} = \frac{V}{\pi \left(\left(\frac{V}{2\pi}\right)^{1/3}\right)^2} = \frac{V}{\pi \left(\frac{V}{2\pi}\right)^{2/3}} = \frac{V}{\pi^{1/3} (\frac{V}{2})^{2/3}} = \frac{V^{1/3} 2^{2/3}}{\pi^{1/3}}\).

   Alternatively, \(h = \frac{V}{\pi r^2} = \frac{2\pi r^3}{\pi r^2} = 2r\).

   So, for minimum surface area, the height \(h\) should be equal to the diameter \(2r\).

This is a classic optimization problem from engineering design. We want to minimize the material used (surface area) for a given volume.

1. Formulate the objective function: We need to minimize \(A\), and it initially depends on \(r\) and \(h\).
2. Introduce constraints: The volume \(V\) is given, providing a constraint equation \(V = \pi r^2 h\).
3. Reduce to a single variable: Use the constraint to eliminate one variable (\(h\)) from the objective function, leaving \(A(r)\).
4. Find critical points: Differentiate \(A(r)\) with respect to \(r\) and find where \(A'(r)=0\) or is undefined. The domain is \((0, \infty)\).
5. Apply First Derivative Test: Analyze the sign of \(A'(r)\) around the critical point to determine if it’s a local minimum. Since the function is decreasing before this point and increasing after, it’s a global minimum on this open interval.
6. Interpret the result: The optimal radius is \(r = (\frac{V}{2\pi})^{1/3}\). Substituting this back into the height equation reveals a famous design principle: for a cylindrical can to minimize surface area for a given volume, its height should be equal to its diameter (\(h=2r\)). This is why many beverage cans have a height roughly twice their radius.

5.10 Second Derivative and the Nature of Extrema

The second derivative provides information about the concavity of a function’s graph.

Definition 5.10.1: Concavity and Inflection Points

1. The graph of \(f\) concaves upward (or is convex) on an interval if it lies above its tangents.
2. The graph of \(f\) concaves downward (or is concave) on an interval if it lies below its tangents.
3. An inflection point \(P\) is a point where \(f\) is continuous and the curve changes concavity (from upward to downward or vice-versa).

Theorem 5.10.2: Second Derivative Test for Concavity

1. If \(f''(x) > 0\) for all \(x\) in \((a, b)\), then the graph of \(f\) concaves upward on \((a, b)\).
2. If \(f''(x) < 0\) for all \(x\) in \((a, b)\), then the graph of \(f\) concaves downward on \((a, b)\).

5.10 Second Derivative and the Nature of Extrema (cont.)

Example 5.10.3: Analyzing Concavity

Let \(f(x) = 2 + 3x - x^3\). Find intervals of concavity and inflection points.

Solution:

1. Find \(f'(x)\): \(f'(x) = 3 - 3x^2\).

2. Find \(f''(x)\): \(f''(x) = -6x\).

3. Analyze the sign of \(f''(x)\):

   • \(f''(x) > 0 \Leftrightarrow -6x > 0 \Leftrightarrow x < 0\).

     So, \(f\) concaves upward on \((-\infty, 0)\).

   • \(f''(x) < 0 \Leftrightarrow -6x < 0 \Leftrightarrow x > 0\).

     So, \(f\) concaves downward on \((0, \infty)\).

4. Inflection Point: At \(x=0\), \(f''(x)\) changes sign, and \(f(x)\) is continuous.

   Therefore, \(x=0\) is an inflection point.

   The point is \((0, f(0)) = (0, 2)\).

The second derivative is a powerful tool for understanding the “shape” or “curvature” of a function’s graph. This is known as concavity.

• A function that “concaves upward” (like a cup holding water) has its graph lying above its tangent lines. Its rate of change (first derivative) is increasing.
• A function that “concaves downward” (like an inverted cup) has its graph lying below its tangent lines. Its rate of change is decreasing.
• An inflection point is where the concavity changes, essentially where the curve switches from bending one way to bending the other.

Theorem 5.10.2 provides the formal link:

• If \(f''(x) > 0\), the function is concave upward.
• If \(f''(x) < 0\), the function is concave downward.
• An inflection point often occurs where \(f''(x) = 0\) or is undefined and the sign of \(f''(x)\) changes.

In Example 5.10.3, we find the first and second derivatives of \(f(x) = 2 + 3x - x^3\). By analyzing the sign of \(f''(x)=-6x\), we determine the intervals of upward and downward concavity. Since \(f''(x)\) changes sign at \(x=0\), this point is an inflection point. This information is vital for accurately sketching graphs and understanding physical phenomena.

5.10.2 The Nature of Extrema (Second Derivative Test)

The second derivative can also help classify local extrema.

Theorem 5.10.4: Global Extrema for Concave Functions

Suppose \(f\) is twice differentiable on \((a, b)\) and \(f'(c) = 0\) for some \(c \in (a, b)\).

1. If \(f''(x) > 0\) on \((a, b)\), then \(f(c)\) is a global minimum on \((a, b)\).
2. If \(f''(x) < 0\) on \((a, b)\), then \(f(c)\) is a global maximum on \((a, b)\).

(This applies when the concavity is uniform over the entire interval.)

5.10.2 The Nature of Extrema (Second Derivative Test) (cont.)

Application to Optimization Problem (Example 5.9.6 revisited):

Minimize surface area of a cylindrical can for a given volume \(V\).

We found \(A(r) = 2\pi r^2 + \frac{2V}{r}\), with domain \((0, \infty)\).

\(A'(r) = 4\pi r - \frac{2V}{r^2}\).

The critical point (stationary point) is \(r = \left(\frac{V}{2\pi}\right)^{1/3}\).

Now, let’s use the Second Derivative Test:

1. Find \(A''(r)\):

   \(A''(r) = \frac{d}{dr}\left(4\pi r - 2Vr^{-2}\right) = 4\pi - 2V(-2)r^{-3} = 4\pi + \frac{4V}{r^3}\).

2. Analyze \(A''(r)\) on \((0, \infty)\):

   Since \(V > 0\) and \(r > 0\), we have \(4\pi > 0\) and \(\frac{4V}{r^3} > 0\).

   Thus, \(A''(r) = 4\pi + \frac{4V}{r^3} > 0\) for all \(r \in (0, \infty)\).

3. Conclusion: Since \(A'(r) = 0\) at \(r = \left(\frac{V}{2\pi}\right)^{1/3}\) and \(A''(r) > 0\) everywhere on the domain, \(A\left(\left(\frac{V}{2\pi}\right)^{1/3}\right)\) is a global minimum.

The second derivative provides a powerful shortcut to classify local extrema, especially when the function is twice differentiable.

Theorem 5.10.4 simplifies optimization when the concavity of the function is consistent over the entire domain. If \(f'(c)=0\) and \(f''(x)>0\) everywhere, then \(f(c)\) is a global minimum. If \(f'(c)=0\) and \(f''(x)<0\) everywhere, then \(f(c)\) is a global maximum. This is a very strong conclusion!

Let’s revisit the cylinder optimization problem. We had already found the critical point \(r = (\frac{V}{2\pi})^{1/3}\) using the first derivative test. Now, we compute the second derivative \(A''(r)\). We observe that \(A''(r)\) is positive for all valid radii. Since \(A''(r) > 0\) means the function is always concave upward, the critical point must be a global minimum. This confirms our earlier conclusion from the first derivative test, often with less analysis of intervals. The second derivative test is often more direct when it applies.

Classifying Local Extrema

When dealing with just local extrema, we use the First or Second Derivative Test.

The First Derivative Test (Theorem 5.10.6):

Suppose \(f\) is continuous in a neighborhood of \(c\) (a critical point) and \(f'\) exists in a deleted neighborhood of \(c\).

1. If \(f'(x)\) changes from negative to positive as \(x\) increases through \(c\), then \(f\) has a local minimum at \(c\).
2. If \(f'(x)\) changes from positive to negative as \(x\) increases through \(c\), then \(f\) has a local maximum at \(c\).
3. If \(f'(x)\) does not change sign as \(x\) increases through \(c\), then \(f\) has no local extremum at \(c\).

Classifying Local Extrema (cont.)

Example 5.10.7: \(f(x) = (x - 1)^{2/3}\)

\(f'(x) = \frac{2}{3}(x - 1)^{-1/3} = \frac{2}{3\sqrt[3]{x-1}}\).

Critical point at \(x=1\) (singular point, \(f'(1)\) undefined).

• For \(x < 1\), \(x-1 < 0 \Rightarrow \sqrt[3]{x-1} < 0 \Rightarrow f'(x) < 0\).
• For \(x > 1\), \(x-1 > 0 \Rightarrow \sqrt[3]{x-1} > 0 \Rightarrow f'(x) > 0\).

Since \(f'(x)\) changes from negative to positive at \(x=1\), \(f(1) = 0\) is a local minimum.

Example 5.10.8: \(f(x) = (x - 1)^{1/3}\)

\(f'(x) = \frac{1}{3}(x - 1)^{-2/3} = \frac{1}{3(x-1)^{2/3}}\).

Critical point at \(x=1\) (singular point, \(f'(1)\) undefined).

• For \(x < 1\), \((x-1)^2 > 0 \Rightarrow (x-1)^{2/3} > 0 \Rightarrow f'(x) > 0\).
• For \(x > 1\), \((x-1)^2 > 0 \Rightarrow (x-1)^{2/3} > 0 \Rightarrow f'(x) > 0\).

Since \(f'(x)\) does not change sign at \(x=1\), \(f(1) = 0\) is neither a local maximum nor a local minimum. (It’s an inflection point with a vertical tangent).

When we’re interested in classifying individual critical points as local maxima, minima, or neither, we have two primary tools: the First Derivative Test and the Second Derivative Test.

The First Derivative Test is very intuitive: it examines the sign of the derivative on either side of a critical point. If the function goes from decreasing to increasing, it’s a local minimum. If it goes from increasing to decreasing, it’s a local maximum. If there’s no change in the derivative’s sign, then it’s neither. This test works even if the derivative doesn’t exist at the critical point (singular points).

Example 5.10.7 (cube root of \((x-1)^2\)) clearly shows a change from negative to positive derivative, indicating a local minimum. This corresponds to a sharp “V” shape at \(x=1\).

Example 5.10.8 (cube root of \((x-1)\)) is an interesting case. The derivative is always positive around \(x=1\). Even though \(f'(1)\) is undefined, the function continues to increase. This tells us there’s no local extremum; it’s an inflection point where the tangent is vertical. These examples highlight the strength and versatility of the First Derivative Test for all types of critical points.

The Second Derivative Test

The Second Derivative Test is often quicker if \(f''(c)\) is easy to compute and non-zero.

Theorem 5.10.9 (Second Derivative Test):

Suppose \(f'(c) = 0\) (so \(c\) is a stationary point) and \(f''\) is continuous near \(c\).

1. If \(f''(c) > 0\), then \(f\) has a local minimum at \(c\).
2. If \(f''(c) < 0\), then \(f\) has a local maximum at \(c\).
3. If \(f''(c) = 0\), the test is inconclusive. (You must use the First Derivative Test or higher derivatives).

Example 5.10.10: Classify critical points of \(f(x) = 2 + 3x - x^3\).

Solution:

\(f'(x) = 3 - 3x^2\). Critical points (where \(f'(x)=0\)) are \(x = 1\) and \(x = -1\).

\(f''(x) = -6x\).

• At \(x = 1\): \(f''(1) = -6(1) = -6\). Since \(f''(1) < 0\), \(f\) has a local maximum at \(x = 1\).
• At \(x = -1\): \(f''(-1) = -6(-1) = 6\). Since \(f''(-1) > 0\), \(f\) has a local minimum at \(x = -1\).

The Second Derivative Test (cont.)

Example 5.10.11: Classify critical points of \(f(x) = x^4\).

Solution:

\(f'(x) = 4x^3\). Critical point (where \(f'(x)=0\)) is \(x = 0\).

\(f''(x) = 12x^2\).

• At \(x = 0\): \(f''(0) = 12(0)^2 = 0\).

  The Second Derivative Test is inconclusive.

We must use the First Derivative Test:

• For \(x < 0\), \(f'(x) = 4x^3 < 0\) (\(f\) is decreasing).
• For \(x > 0\), \(f'(x) = 4x^3 > 0\) (\(f\) is increasing).

Since \(f'(x)\) changes from negative to positive at \(x=0\), \(f(0)=0\) is a local minimum.

The Second Derivative Test is a powerful method for classifying stationary points (where \(f'(c)=0\)). It’s often quicker than the First Derivative Test, but it has limitations.

The test works by looking at the concavity at the critical point:

• If \(f''(c) > 0\), the function is concave up, so \(c\) is a local minimum.
• If \(f''(c) < 0\), the function is concave down, so \(c\) is a local maximum.

However, if \(f''(c)=0\), the test is inconclusive. This means you can’t determine the nature of the critical point from the second derivative alone. In such cases, you must revert to the First Derivative Test or use higher-order derivatives (which is beyond this scope for now).

Example 5.10.10 shows a successful application: for \(f(x) = 2 + 3x - x^3\), the second derivative quickly classifies \(x=1\) as a local max and \(x=-1\) as a local min.

Example 5.10.11 (\(f(x)=x^4\)) illustrates the inconclusive case. At \(x=0\), \(f'(0)=0\) and \(f''(0)=0\). We then switch to the First Derivative Test, which confirms that \(x=0\) is indeed a local minimum. This example highlights the importance of knowing both tests and when to use each.

5.10.3 Curve Sketching

We can combine all the information from derivatives to accurately sketch the graph of a function.

Useful Steps for Curve Sketching:

1. Intervals of Increase/Decrease: Find \(f'(x)\) and determine where \(f'(x) > 0\) (increasing) and \(f'(x) < 0\) (decreasing).
2. Local Extrema: Use the First or Second Derivative Test to classify critical points.
3. Intervals of Concavity: Find \(f''(x)\) and determine where \(f''(x) > 0\) (concave up) and \(f''(x) < 0\) (concave down).
4. Inflection Points: Identify points where concavity changes (and \(f\) is continuous).
5. Vertical Asymptotes: Find \(a\) where \(\lim_{x \to a^+} f(x) = \pm \infty\) or \(\lim_{x \to a^-} f(x) = \pm \infty\).
6. Horizontal Asymptotes: Find \(b\) where \(\lim_{x \to \infty} f(x) = b\) or \(\lim_{x \to -\infty} f(x) = b\).
7. Intercepts: Find \(x\)-intercepts (\(f(x)=0\)) and \(y\)-intercept (\(f(0)\)).
8. Plot and Sketch: Use all this information to accurately sketch the graph.

Curve sketching is the grand finale of differentiation. It’s where all the concepts—first derivatives, second derivatives, limits, and continuity—come together to allow us to visualize a function’s behavior without simply relying on a graphing calculator.

The steps outline a systematic approach:

• The first derivative tells us about the function’s direction (uphill/downhill) and helps identify local peaks and valleys.
• The second derivative tells us about the function’s curvature (how it bends) and helps identify inflection points where the bending changes.
• Asymptotes tell us about the function’s behavior at the edges of its domain or as x goes to infinity.
• Intercepts provide anchor points on the axes.

By gathering all this information, you can piece together a remarkably accurate sketch of a function, revealing its essential characteristics. This process is not just about drawing a picture; it’s about deeply understanding the function’s mathematical properties.

Example: Sketching \(y = 2 + 3x - x^3\)

Example 5.10.12: Sketch the graph of \(y = 2 + 3x - x^3\).

Solution: Let \(f(x) = 2 + 3x - x^3\).

1. Derivatives:

   • \(f'(x) = 3 - 3x^2 = 3(1 - x)(1 + x)\).
   • \(f''(x) = -6x\).

2. Intervals of Increase/Decrease:

   • \(f'(x) > 0\) for \(x \in (-1, 1)\) (\(f\) is increasing).
   • \(f'(x) < 0\) for \(x \in (-\infty, -1) \cup (1, \infty)\) (\(f\) is decreasing).

3. Local Extrema (using Second Derivative Test):

   Critical points: \(f'(x)=0 \Rightarrow x=1, x=-1\).

   • At \(x=1\): \(f''(1) = -6(1) = -6 < 0\). Local maximum at \((1, f(1)) = (1, 4)\).
   • At \(x=-1\): \(f''(-1) = -6(-1) = 6 > 0\). Local minimum at \((-1, f(-1)) = (-1, 0)\).

Example: Sketching \(y = 2 + 3x - x^3\)

4. Intervals of Concavity:

   • \(f''(x) > 0\) for \(x < 0\) (\(f\) is concave upward on \((-\infty, 0)\)).
   • \(f''(x) < 0\) for \(x > 0\) (\(f\) is concave downward on \((0, \infty)\)).

5. Inflection Point:

   At \(x=0\), \(f''(x)\) changes sign. \(f(0) = 2\).

   Inflection point at \((0, 2)\).

6. Asymptotes:

   • No vertical asymptotes (polynomial is continuous everywhere).
   • No horizontal asymptotes: \(\lim_{x\to\infty} f(x) = -\infty\), \(\lim_{x\to-\infty} f(x) = \infty\).

7. Intercepts:

   • \(y\)-intercept: \(f(0) = 2\). Point \((0, 2)\).
   • \(x\)-intercepts: \(2 + 3x - x^3 = 0\). (Hard to find exactly, but the plot will reveal it).

Sketch:

Figure: Sketch of the graph y = 2 + 3x - x^3, showing local extrema, inflection point, and concavity.

Let’s put all our curve sketching tools to use for the function \(f(x) = 2 + 3x - x^3\).

1. Derivatives: We start by finding the first and second derivatives.
2. Increasing/Decreasing: Analyze the sign of \(f'(x)\) to determine where the function rises and falls. We find \(f\) increases on \([-1,1]\) and decreases elsewhere.
3. Local Extrema: Using the critical points from \(f'(x)=0\) and the Second Derivative Test, we identify a local maximum at \(x=1\) and a local minimum at \(x=-1\). We also compute their \(y\)-values.
4. Concavity: Analyze the sign of \(f''(x)\) to determine concavity. We find concave up for \(x<0\) and concave down for \(x>0\).
5. Inflection Point: Since concavity changes at \(x=0\), and the function is continuous, \((0,2)\) is an inflection point.
6. Asymptotes: As a polynomial, there are no vertical or horizontal asymptotes. The function tends to \(\pm \infty\) at the extremes.
7. Intercepts: We find the \(y\)-intercept. While \(x\)-intercepts can be hard for cubics, we know there must be one based on the local extrema and end behavior.

Finally, we combine all this information to draw the graph. We plot the critical points, inflection points, and intercepts. Then, we connect these points, making sure the curve reflects the correct increasing/decreasing behavior and concavity in each interval. This systematic approach allows for a very accurate sketch.

Key Takeaways

Differentiation is the study of change, crucial for understanding and modeling dynamic systems.

• Derivatives as Slope & Rate of Change: The derivative \(f'(x)\) represents the slope of the tangent line to \(f(x)\) and the instantaneous rate of change of \(f\) with respect to \(x\).
• Differentiability & Continuity: Differentiability implies continuity, but not vice-versa. Functions with kinks or breaks are not differentiable.
• Core Differentiation Rules:
  • Power Rule: \(\frac{d}{dx}(x^n) = nx^{n-1}\) (for all real \(n\)).
  • Constant Multiple Rule: \(\frac{d}{dx}(cf(x)) = cf'(x)\).
  • Sum/Difference Rule: \(\frac{d}{dx}(f \pm g) = f' \pm g'\).
  • Product Rule: \(\frac{d}{dx}(fg) = f'g + fg'\).
  • Quotient Rule: \(\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2}\).
  • Chain Rule: \(\frac{d}{dx}(f(g(x))) = f'(g(x))g'(x)\).

Key Takeaways

• Derivatives of Common Functions: Memorize derivatives of trig, exponential, and logarithmic functions.
• Implicit & Parametric Differentiation: Techniques for finding derivatives when functions are not explicitly defined in terms of \(x\).
• Linearization & Differentials: Approximating functions and estimating changes using tangent lines.
• Newton’s Method: An iterative numerical technique for finding roots of equations using derivatives.
• Optimization: Finding global/local maxima and minima using critical points and the First/Second Derivative Tests.
• Concavity & Inflection Points: Using the second derivative to determine the curvature of a graph.

Let’s recap the most important concepts from this chapter.

Differentiation is fundamentally about understanding change, both geometrically as the slope of a tangent and physically as an instantaneous rate of change.

We covered the critical relationship between differentiability and continuity.

The core of practical differentiation lies in mastering the fundamental rules: the power rule, constant multiple, sum/difference, product, quotient, and especially the chain rule. You also need to know the derivatives of the standard trigonometric, exponential, and logarithmic functions.

Beyond these basic computations, we explored more advanced techniques like implicit and parametric differentiation, which expand the types of functions we can analyze.

We then saw the powerful applications of derivatives: - Linearization and differentials for approximation and error estimation. - Newton’s Method for numerical root finding. - Optimization for finding maximums and minimums, guided by critical points and derivative tests. - Concavity and inflection points, which use the second derivative to understand the shape of a graph.

These tools are indispensable in engineering and mathematics.

Key Equations

Equation	Description
\(f ^ {\prime} (x) = \lim _ {h \to 0} \frac {f (x + h) - f (x)}{h}\)	Definition of the derivative
\(\frac {d}{d x} (C) = 0\)	Derivative of a constant
\(\frac {d}{d x} (x ^ {n}) = n x ^ {n - 1}\)	Power Rule (for any real \(n\))
\(\frac {d}{d x} (c f (x)) = c f ^ {\prime} (x)\)	Constant Multiple Rule
\(\frac {d}{d x} (f (x) \pm g (x)) = f ^ {\prime} (x) \pm g ^ {\prime} (x)\)	Sum/Difference Rule
\(\frac {d}{d x} (f (x) g (x)) = f ^ {\prime} (x) g (x) + f (x) g ^ {\prime} (x)\)	Product Rule
\(\frac {d}{d x} \left(\frac {f (x)}{g (x)}\right) = \frac {f ^ {\prime} (x) g (x) - f (x) g ^ {\prime} (x)}{(g (x)) ^ {2}}\)	Quotient Rule
\(\frac {d}{d x} (f (g (x))) = f ^ {\prime} (g (x)) g ^ {\prime} (x)\)	Chain Rule
\(\frac {d y}{d x} = \frac {\frac {d y}{d t}}{\frac {d x}{d t}}\)	Parametric Differentiation

Key Equations

Equation	Description
\(L (x) = f (a) + f ^ {\prime} (a) (x - a)\)	Linearization of \(f\) at \(a\)
\(x _ {n + 1} = x _ {n} - \frac {f (x _ {n})}{f ^ {\prime} (x _ {n})}\)	Newton’s Method iteration formula
\(\lim _ {x \to a} \frac {f (x)}{g (x)} = \lim _ {x \to a} \frac {f ^ {\prime} (x)}{g ^ {\prime} (x)}\)	L’Hospital’s Rule (for indeterminate forms)
If \(f'(x) > 0 \implies f\) is increasing; if \(f'(x) < 0 \implies f\) is decreasing	First Derivative Test for Monotonicity
If \(f''(c) > 0 \implies\) local min; if \(f''(c) < 0 \implies\) local max	Second Derivative Test for Extrema (at stationary points)
If \(f''(x) > 0 \implies f\) is concave up; if \(f''(x) < 0 \implies f\) is concave down	Second Derivative Test for Concavity

This table summarizes the essential equations from the chapter. These are the formulas and definitions you’ll be using constantly in your math and engineering courses. Make sure you understand each one and can apply it correctly. The first few are definitions, followed by the fundamental rules of differentiation. Then we have techniques for specific scenarios like parametric and implicit differentiation. Finally, we have the application-oriented formulas for linearization, Newton’s method, and L’Hospital’s rule, along with the tests for function behavior.

Key Terms

Term	Definition
Derivative	The instantaneous rate of change of a function with respect to its independent variable; the slope of the tangent line to the function’s graph at a point.
Differentiable	A function is differentiable at a point if its derivative exists (i.e., the limit defining the derivative is finite) at that point.
Tangent Line	A straight line that “just touches” a curve at a single point, representing the local direction of the curve.
Instantaneous Rate of Change	The rate at which a quantity changes at a particular moment in time or at a specific point.
Higher Derivatives	Derivatives obtained by differentiating a function multiple times (e.g., second derivative, third derivative).
Implicit Differentiation	A technique used to find the derivative of a function defined by an equation that implicitly relates variables, rather than explicitly.

Key Terms

Term	Definition
Parametric Differentiation	A technique used to find \(\frac{dy}{dx}\) when \(x\) and \(y\) are both expressed as functions of a third parameter (e.g., \(t\)).
Logarithmic Differentiation	A method of differentiating functions by first taking the natural logarithm of both sides and then differentiating implicitly.
Linearization	The process of approximating a function near a point using its tangent line at that point.
Differentials	An approximation of the actual change in a function (\(\Delta f\)) using the derivative and a small change in the independent variable (\(df = f'(x)dx\)).
Newton’s Method	An iterative numerical method for approximating the roots (zeros) of a real-valued function using its tangent lines.
Optimization	The mathematical process of finding the maximum or minimum values of a function, often subject to constraints.
Critical Point	A point in the domain of a function where its derivative is either zero (stationary point) or undefined (singular point).

Key Terms

Term	Definition
Local Maximum/Minimum	A point where the function’s value is greater/less than all nearby values in its domain.
Global Maximum/Minimum	The absolute largest/smallest value of a function over its entire domain.
Concave Upward/Downward	Describes the curvature of a function’s graph. Concave upward means the graph bends like a “U”; concave downward means it bends like an “inverted U”.
Inflection Point	A point on the graph of a function where the concavity changes (from upward to downward or vice-versa).
L’Hospital’s Rule	A theorem used to evaluate limits of indeterminate forms (like \(\frac{0}{0}\) or \(\frac{\pm\infty}{\pm\infty}\)) by taking derivatives of the numerator and denominator.
Mean Value Theorem	States that for a continuous and differentiable function on an interval, there is at least one point where the instantaneous rate of change equals the average rate of change.

This glossary provides definitions for the key terms introduced in this chapter. Understanding these terms is crucial for mastering differentiation. Pay special attention to the nuanced definitions, like the difference between a critical point and a local extremum, or concave up vs. concave down. These terms form the vocabulary you’ll need to discuss and apply calculus concepts effectively in your studies and future career.