Calculus I: Foundations

Continuity and the Intermediate Value Theorem

Imron Rosyadi

Calculus I: Foundations

Continuity and the Intermediate Value Theorem

Welcome everyone! Today, we’re diving into two foundational concepts in calculus: Continuity and the Intermediate Value Theorem. These ideas are absolutely crucial for understanding how functions behave, and they lay the groundwork for differentiation and integration, which we’ll explore later.

What is Continuity?

An Intuitive Understanding

Imagine drawing a function’s graph without lifting your pencil.

Continuous Function

• No breaks.
• No jumps.
• No holes.
• Smooth transitions.

Discontinuous Function

• Gaps or breaks.
• Sudden jumps.
• “Holes” or missing points.

Tip

Think of it like a smooth road: A continuous function is a road you can drive on without hitting any unexpected cliffs, bridges, or missing sections!

Formal Definition of Continuity at a Point

When is a function continuous at \(x=a\)?

A function \(f(x)\) is continuous at \(a\) if all three conditions below are met:

1. \(f(a)\) is defined.
2. \(\displaystyle\lim_{x \to a} f(x)\) exists.
3. \(\displaystyle\lim_{x \to a} f(x) = f(a)\).

In other words, the left-hand limit, the function value, and the right-hand limit must all be equal: \[ \lim_{x \to a^-} f(x)\qquad=\qquad f(a)\qquad=\qquad\lim_{x \to a^+} f(x) \]

Important

If any of these conditions fail, \(f(x)\) is discontinuous at \(x=a\).

The intuitive idea of drawing without lifting a pencil translates directly to these three formal conditions. For a function to be continuous at a point ‘a’, first, the function must actually have a value at ‘a’. Second, as you approach ‘a’ from both the left and the right, the function should be heading towards a single, predictable value – meaning the limit exists. And finally, that predictable value must be exactly what the function is at ‘a’. If any of these don’t hold true, we have a discontinuity.

Visualizing Continuity Conditions

How the three conditions fit together logically.

graph LR
    A["Is f(a) 
    defined?"] -->|Yes| B
    A -->|No| D{Discontinuous}
    B["Does lim_x->a f(x) 
    exist?"] -->|Yes| C
    B -->|No| D
    C["Is lim_x->a f(x) 
    == f(a)?"] -->|Yes| E{Continuous 
    at 'a'}
    C -->|No| D

This Mermaid diagram provides a visual flowchart for checking the three conditions of continuity at a point ‘a’. Starting from the top, you sequentially check each condition. Only if all three are met does the function qualify as continuous at that specific point. If any condition fails, the function is discontinuous.

Graphical Interpretation & Your First “DO”

Graphically: Imagine tracing \(f(x)\) with a pencil. To be continuous, you shouldn’t lift your pencil as you pass through \(x=a\).

Your Turn: Sketch \(f(x)=\sqrt x\) and let \(a=4\).

Exercise:

1. Find \(f(a)\).
2. Find \(\displaystyle\lim_{x\to a^-}f(x)\).
3. Find \(\displaystyle\lim_{x\to a^+}f(x)\).
4. Is \(f(x)=\sqrt x\) continuous at \(x=a\)?

Think:

• Does \(f(4)\) exist?
• What happens as \(x\) approaches 4 from the left?
• What happens as \(x\) approaches 4 from the right?
• Do these values match \(f(4)\)?

This exercise helps solidify the conceptual understanding. Let’s walk through it for \(f(x) = \sqrt{x}\) at \(a=4\): First, \(f(4) = \sqrt{4} = 2\). Second, the limit as \(x\) approaches 4 from the left, \(\lim_{x \to 4^-}\sqrt{x}\), is 2. Third, the limit as \(x\) approaches 4 from the right, \(\lim_{x \to 4^+}\sqrt{x}\), is also 2. Since all three values are equal to 2, the function is continuous at \(x=4\). Although \(\sqrt{x}\) is only defined for \(x \ge 0\), for \(x=4\), all continuity conditions are met. We’ll discuss continuity on intervals shortly, which clarifies definitions at domain boundaries.

When Continuity Fails: An Example

The case of division by zero.

Consider the function \(f(x) = \frac{x^2-1}{x-1}\).

Question: Where might this function be discontinuous?

Let’s analyze it:

1. Domain: \(x \ne 1\). So \(f(1)\) is undefined.
2. Simplification: For \(x \ne 1\), \(f(x) = \frac{(x-1)(x+1)}{x-1} = x+1\).

This function has a hole at \(x=1\).

This YouTube video provides a visual illustration of this common type of discontinuity. Because the function \(f(x)\) is undefined at \(x=1\), the first condition for continuity fails, making it discontinuous. If we were to redefine \(f(1)\) to be the limit value, which is 2, then the function would become continuous. This specific type of discontinuity is called a removable discontinuity, which we’ll discuss next.

Three Main Types of Discontinuities

Ways a function can fail to be continuous.

1. Jump Discontinuity

• Limits from left and right exist but are not equal.
• A sudden “jump” in the graph.

Jump Discontinuity

2. Infinite Discontinuity

• One or both limits approach \(\pm \infty\).
• Often involves vertical asymptotes.

Infinite Discontinuity

These images clearly illustrate two of the three primary ways functions can be discontinuous. A jump discontinuity typically occurs in piecewise functions where the pieces don’t meet up at the point of change. An infinite discontinuity happens when the function’s value shoots off to positive or negative infinity, often associated with division by zero or vertical asymptotes.

Three Main Types of Discontinuities (cont.)

Ways a function can fail to be continuous.

3. Removable Discontinuity

• Limits from left and right are equal and finite, but \(f(a)\) either:
  • Is undefined, OR
  • Is defined but equals a different value.
• A “hole” in the graph, which can be “filled” by redefining \(f(a)\).

Removable Discontinuity

The third type, removable discontinuity, is special because it means the function almost behaves continuously at a point; the limit exists. The issue is either that the function isn’t defined there, or it’s defined but at the “wrong” value. We call it removable because you can mathematically patch the hole by redefining the function at that single point. The (x^2-1)/(x-1) example from earlier is a classic case of a removable discontinuity.

Continuity on an Interval

Beyond a single point.

A function \(f\) is continuous on an interval \(I\) if it is continuous at:

1. Each interior point of \(I\).
2. The left endpoint (if \(I\) has one), it must be continuous from the right: \(\displaystyle \lim_{x \rightarrow a^+} f(x) = f(a)\).
3. The right endpoint (if \(I\) has one), it must be continuous from the left: \(\displaystyle \lim_{x \rightarrow a^-} f(x) = f(a)\).

Note

One-sided continuity is essential for discussing continuity at the endpoints of closed or half-open intervals.

This slide extends our concept of continuity beyond just a single point to an entire interval. For points within the interval, the standard three conditions apply. However, at the very beginning or end of a closed interval, we can only approach the endpoint from one side. This is where one-sided limits and one-sided continuity become crucial. The video provides further explanation and examples of these concepts.

Continuity and Piecewise Functions

Checking continuity where the definition changes.

If a function’s definition changes at \(x=a\), we must carefully compare the three continuity conditions at that specific point.

Example: Consider \(f(x) = \begin{cases} 1-x, & x < 0 \\ x^2, & x \ge 0 \end{cases}\)

Let’s check continuity at \(x=0\):

1. Left-hand limit: \(\displaystyle\lim_{x \to 0^-} f(x)= \lim_{x \to 0^-} (1-x) = 1\)
2. Right-hand limit: \(\displaystyle\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2) = 0\)
3. Function value: \(f(0) = 0^2 = 0\)

Since \(\displaystyle\lim_{x \to 0^-} f(x) \ne \lim_{x \to 0^+} f(x)\), the limit \(\displaystyle\lim_{x \to 0} f(x)\) does not exist. Therefore, \(f(x)\) is discontinuous at \(x=0\) (a jump discontinuity).

Piecewise functions are excellent tools for practicing the continuity conditions, especially at the points where the function’s definition changes. In this example, we clearly see that the left-hand limit approaches 1, while the right-hand limit and the function value approach 0. Since the left and right limits don’t match, the overall limit at \(x=0\) does not exist, leading to a jump discontinuity. Always remember to use the correct piece of the function for the corresponding interval when evaluating limits.

The Intermediate Value Theorem (IVT)

Guaranteering existence of values.

The Intermediate Value Theorem (IVT) states:

Suppose \(f(x)\) is a continuous function on the closed interval \([a,b]\) with \(f(a) \ne f(b)\). If \(N\) is any number between \(f(a)\) and \(f(b)\), then there exists at least one point \(c\) in the open interval \((a,b)\) such that \(f(c)=N\).

In simpler terms:

If you draw a continuous curve from point A to point B, you must pass through every y-value between the y-value of A and the y-value of B.

The IVT is incredibly powerful because it guarantees the existence of a solution or a specific function value, even if we can’t find it explicitly. The absolute key condition for the IVT to apply is continuity. If the function has a jump, a hole, or an asymptote, the theorem does not necessarily hold. The video provides a great graphical explanation that brings this abstract concept to life.

IVT Application: Proving the Existence of Roots

A powerful tool for complex equations.

We can use the IVT to show that certain equations have solutions or that polynomials have roots.

Example:

Show that \(f(x)=x^4+x-3\) has a root between \(x=-2\) and \(x=0\).

1. Check Continuity: \(f(x)\) is a polynomial, so it is continuous everywhere.
2. Evaluate Endpoints:
   • \(f(-2) = (-2)^4 + (-2) - 3 = 16 - 2 - 3 = 11\)
   • \(f(0) = (0)^4 + (0) - 3 = -3\)
3. Choose N: We are looking for a root, so \(N=0\).
   • Notice that \(f(0) = -3 < 0 < f(-2) = 11\). So \(N=0\) is indeed between \(f(-2)\) and \(f(0)\).
4. Conclusion by IVT: Since \(f(x)\) is continuous on the interval \([-2,0]\) and \(0\) is between \(f(-2)\) and \(f(0)\), the Intermediate Value Theorem guarantees there exists at least one \(c \in (-2,0)\) such that \(f(c)=0\).

This means \(f(x)\) has a root in the interval \((-2,0)\).

Tip

We don’t need to find the exact root, just prove its existence!

This example demonstrates the practical usefulness of the IVT. Even for a polynomial that might be difficult to factor or solve analytically, we can use simple evaluations to quickly confirm that a root must exist within a given interval. This principle is fundamental to many numerical methods used in computing and engineering for finding approximate solutions.

Case Study: IVT in Action - Temperature Gradient

A practical application in environmental science.

Imagine a sensor measuring ground temperature at 1 meter depth at 9 AM on two consecutive days.

Scenario:

• Day 1 (9 AM): Temperature at 1 meter depth: \(10^\circ C\).
• Day 2 (9 AM): Temperature at 1 meter depth: \(15^\circ C\).

Assume temperature changes continuously over time.

Question: Is it guaranteed that there was a moment between Day 1 and Day 2 when the ground temperature at 1 meter depth was exactly \(12^\circ C\)?

Important

Yes, due to IVT!

• Let \(T(t)\) be the temperature at time \(t\).
• We assume \(T(t)\) is continuous over the time interval.
• \(T(\text{Day 1}) = 10^\circ C\), \(T(\text{Day 2}) = 15^\circ C\).
• \(N = 12^\circ C\) is a value between \(10^\circ C\) and \(15^\circ C\).
• Therefore, by the IVT, there must have been a time \(c\) between Day 1 and Day 2 where \(T(c) = 12^\circ C\).

This real-world example helps students connect the abstract IVT to a tangible situation. It illustrates that if a physical quantity, like temperature, varies continuously over an interval, it must pass through all intermediate values. This principle applies in many scientific and engineering fields, such as analyzing fluid flow, structural stress, or even financial market changes, whenever continuous models are appropriate.

Practical IVT: Finding Roots Numerically

Using Python (Pyodide) for interval bisection.

While IVT doesn’t give the exact root, it can help narrow down its location. The bisection method is a numerical technique that iteratively uses the IVT.

Let’s use Pyodide to check an interval for a potential root of \(f(x) = x^3 - 2x - 5\).

This Pyodide block demonstrates a simple application of the IVT principle in numerical methods. The core idea is that if a continuous function changes sign over an interval, a root must lie within that interval. The bisection method repeatedly uses this sign-change logic to shrink the interval containing a root, providing increasingly precise approximations. It illustrates that even in the absence of an exact analytical solution, calculus provides tools for approximation, which is fundamental in computational science.

Properties of Continuous Functions

Building more complex continuous functions.

If \(f\) and \(g\) are both continuous on an interval \(I\), then the following are also continuous on \(I\):

1. Sum: \(f(x) + g(x)\).
2. Difference: \(f(x) - g(x)\).
3. Product: \(f(x) \cdot g(x)\).
4. Quotient: \(\frac{f(x)}{g(x)}\), provided \(g(x) \ne 0\) on \(I\).
5. Composition: If \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a)\), then \(f(g(x))\) is continuous at \(a\).

Tip

These properties allow us to determine the continuity of complex functions by examining their simpler, continuous components.

These theorems are incredibly useful because they mean we don’t have to go back to the formal definition of continuity for every single function we encounter. If we know that the basic building blocks of a function, like polynomials or trigonometric functions, are continuous, then we can confidently deduce the continuity of their sums, differences, products, quotients (being careful about division by zero), and even compositions within their domains. This simplifies our analysis considerably.

Inherently Continuous Functions

The building blocks of continuity.

The following types of functions are continuous on their respective domains:

• Polynomials: E.g., \(f(x) = x^3 - 5x + 7\).
• Rational Functions: E.g., \(f(x) = \frac{x+1}{x-2}\) (discontinuous only where denominator is zero).
• Root Functions: E.g., \(f(x) = \sqrt{x}\) (continuous on \([0, \infty)\)).
• Trigonometric Functions: E.g., \(\sin(x), \cos(x)\) (continuous everywhere), \(\tan(x)\) (continuous on its domain).
• Inverse Trigonometric Functions: E.g., \(\arcsin(x)\).
• Exponential Functions: E.g., \(e^x, 2^x\).
• Logarithmic Functions: E.g., \(\ln(x), \log_{10}(x)\) (continuous on their domains).

Note

Crucial Implication for Limits: If \(f(x)\) is continuous at \(x=a\), then \(\displaystyle \lim_{x\to a} f(x) = f(a)\). This means for continuous functions, you can find the limit by direct substitution!

This slide lists common functions that students frequently encounter and should recognize as inherently continuous within their domains. The most vital takeaway here is the implication for limits. For any function that is continuous at a specific point ‘a’, finding the limit as x approaches ‘a’ is as simple as directly substituting ‘a’ into the function. This incredibly simplifies limit evaluation for a large class of functions.

Summary & Key Takeaways

• Continuity at a Point: Three conditions must be met: \(f(a)\) defined, limit exists, limit equals \(f(a)\).
• Types of Discontinuities: Jump, Infinite, Removable.
• Continuity on Intervals: Defined for interior points and one-sided at endpoints.
• Piecewise Functions: Check carefully at the “split” points.
• Intermediate Value Theorem (IVT):
  • Guarantees the existence of intermediate values for continuous functions on a closed interval.
  • Powerful for proving existence of roots/solutions.
• Properties of Continuous Functions: Sums, differences, products, quotients (if denom \(\ne 0\)), and compositions of continuous functions are also continuous.
• Evaluate Limits by Substitution: For continuous functions, \(\lim_{x\to a} f(x) = f(a)\).

This summary slide is designed to consolidate the main concepts we’ve covered today. It serves as a quick recap and highlights the most important points for you to remember as you continue your studies in calculus. Continuity is the bedrock for many advanced topics, so a solid understanding is key.