Confirming Continuity over an Interval - AP Calculus AB Study Guide

The Core Idea: Confirming Continuity over an Interval

In calculus, we often need to know if a function is "well-behaved" over a specific stretch of its graph, meaning it has no breaks, jumps, or holes. This property is called continuity. Instead of testing every single point within an interval, which is impossible, this topic provides a powerful framework for confirming continuity. The core idea is that we can build complex continuous functions from simpler ones whose continuity is already known.

By understanding that basic functions like polynomials, trigonometric, and exponential functions are continuous everywhere they are defined (their domains), we can use rules about their sums, differences, products, quotients, and compositions to determine the intervals over which more complex functions are continuous. Essentially, the process of confirming continuity over an interval boils down to finding the function's domain.

Key Rules & Definitions

Definition: Continuity on an Interval

A function $f(x)$ is continuous on an interval if it is continuous at every point within that interval.

Continuity of Foundational Functions

The following types of functions are continuous at all points in their natural domains. The key is to know what those domains are.

• Polynomial Functions: Continuous on $(-\infty ,\infty )$. Example: $f(x)=x^3-2x+5$.

• Rational Functions: Continuous everywhere except where the denominator is zero. Example: $f(x)=\frac{x+1}{x-3}$ is continuous on $(-\infty ,3)\cup (3,\infty )$.

• Power Functions: Continuous on their domains. Example: $f(x)=x^{2/3}$ is continuous on $(-\infty ,\infty )$, while $f(x)=x^{-1}=\frac{1}{x}$ is continuous on $(-\infty ,0)\cup (0,\infty )$.

• Exponential Functions: Continuous on $(-\infty ,\infty )$. Example: $f(x)=e^x$ or $f(x)=5^x$.

• Logarithmic Functions: Continuous on $(0,\infty )$. Example: $f(x)=\ln (x)$.

• Trigonometric Functions: Continuous on their domains. For example, $f(x)=\sin (x)$ and $f(x)=\cos (x)$ are continuous on $(-\infty ,\infty )$. However, $f(x)=\tan (x)$ is continuous only on intervals where $\cos (x)\ne 0$, such as $(-\frac{\pi }{2},\frac{\pi }{2})$.

Properties of Continuous Functions

If $f(x)$ and $g(x)$ are continuous functions, then the following combinations are also continuous on their respective domains:

• Sum/Difference:$f(x)\pm g(x)$

• Product:$f(x)\cdot g(x)$

• Quotient:$\frac{f(x)}{g(x)}$ (provided $g(x)\ne 0$)

• Composition:$f(g(x))$ (The domain depends on the domains of both $f$ and $g$)

Understanding Domains

The Essential Knowledge for this topic repeatedly states that functions are continuous "on their domains." Therefore, the most critical skill for confirming continuity is correctly identifying a function's domain. A function is continuous everywhere it is mathematically defined. The task is to find any and all $x$-values that would cause the function to be undefined.

The primary sources of discontinuities arise from three main situations:

1. Division by Zero: For any rational expression $\frac{f(x)}{g(x)}$, you must find all $x$-values that make the denominator $g(x)=0$. These values are not in the domain and are points of discontinuity.

2. Even Roots of Negative Numbers: For any function involving an even root, such as $\sqrt{f(x)}$ or $(f(x){)}^{1/4}$, the expression inside the root (the radicand) must be non-negative. You must solve the inequality $f(x)\ge 0$. The function is discontinuous where $f(x)<0$.

3. Logarithms of Non-Positive Numbers: For any function involving a logarithm, such as $\ln (f(x))$, the argument of the log must be strictly positive. You must solve the inequality $f(x)>0$. The function is discontinuous where $f(x)\le 0$.

Core Concepts & Rules

• A function is continuous across an interval if its graph can be drawn without lifting your pen. Formally, it must be continuous at every individual point in the interval.

• Common functions (polynomial, exponential, sine, cosine) are continuous everywhere. Other foundational functions (logarithmic, rational, other trig functions) are continuous everywhere they are defined.

• You can build complex continuous functions by adding, subtracting, multiplying, dividing, or composing simpler continuous functions.

• The continuity of a combined function holds on its new, potentially more restrictive, domain.

• When analyzing a function for continuity, your primary goal is to determine its domain by looking for potential division by zero, even roots of negative numbers, or logarithms of non-positive numbers.

Step-by-Step Example 1: Rational Function

Problem: Determine the interval(s) over which the function $f(x)=\frac{x^2+5x}{x^2-x-6}$ is continuous.

Step 1: Identify the component functions.

The numerator, $n(x)=x^2+5x$, is a polynomial. The denominator, $d(x)=x^2-x-6$, is also a polynomial. Polynomials are continuous everywhere, i.e., on $(-\infty ,\infty )$.

Step 2: Apply the rule for quotients of continuous functions.

The function $f(x)$ is a quotient of two continuous functions. Therefore, $f(x)$ will be continuous everywhere except for the values of $x$ that make the denominator equal to zero.

Step 3: Find the roots of the denominator.

Set the denominator equal to zero and solve for $x$:

$$x^2-x-6=0$$

$$(x-3)(x+2)=0$$

The roots are $x=3$ and $x=-2$. These are the points of discontinuity.

Step 4: State the intervals of continuity.

The function is continuous for all real numbers except $x=3$ and $x=-2$. We write this in interval notation.

Answer: The function $f(x)$ is continuous on $(-\infty ,-2)\cup (-2,3)\cup (3,\infty )$.

Step-by-Step Example 2: Composition with a Logarithm

Problem: Determine the interval(s) over which the function $h(x)=\ln (4-x^2)$ is continuous.

Step 1: Deconstruct the composite function.

The function $h(x)$ is a composition of $f(u)=\ln (u)$ (the outer function) and $g(x)=4-x^2$ (the inner function).

Step 2: Analyze the continuity of the inner and outer functions.

• The inner function, $g(x)=4-x^2$, is a polynomial and is continuous for all real numbers.

• The outer function, $f(u)=\ln (u)$, is a logarithmic function. It is continuous on its domain, which is $u>0$.

Step 3: Apply the condition for the domain of the composite function.

For $h(x)=\ln (4-x^2)$ to be continuous, its argument must be positive. The continuity of $h(x)$ is restricted by the domain requirement of the natural logarithm. We must ensure the "inside" part, $4-x^2$, is strictly greater than zero.

Step 4: Solve the inequality.

$$4-x^2>0$$

$$4>x^2$$

$$x^2<4$$

Taking the square root of both sides, we get $\sqrt{x^2}<\sqrt{4}$, which means $|x|<2$. This inequality is equivalent to $-2<x<2$.

Step 5: State the interval of continuity.

The function $h(x)$ is defined and continuous only on the interval where $4-x^2>0$.

Answer: The function $h(x)$ is continuous on the open interval $(-2,2)$.

Using Your Calculator

This topic is primarily analytical, meaning you are expected to determine continuity by examining the function's structure and domain rules. A calculator is best used as a tool to verify your conclusion visually.

To verify your answer:

1. Press the [Y=]` button and enter the function. For Example 2, you would enter `Y1 = ln(4 - X^2)`. 2. Press the `[GRAPH]` button. 3. Observe the graph. You should see that the graph for $h(x) = \ln(4-x^2) only exists between $x=-2$ and $x=2$, confirming your analytical result. For a rational function like in Example 1, you would see vertical asymptotes at $x=-2$ and $x=3$, which are visual representations of the discontinuities.

There is no specific calculator function that will automatically find and state the intervals of continuity for you. You must use the rules of continuity first and then use the graph to check your work.

AP Exam Quick Hit

Common Question Types

• Identifying Intervals from an Equation: This is the most common type. "Find the interval(s) on which $f(x)=\frac{\sqrt{x+1}}{x-1}$ is continuous." You must combine the rule for square roots ($x+1\ge 0⟹x\ge -1$) and the rule for quotients ($x-1\ne 0⟹x\ne 1$) to get the final answer: $[-1,1)\cup (1,\infty )$.

• Justifying Conditions for Theorems: "Explain why the Intermediate Value Theorem can be applied to the function $f(x)=e^x-\cos (x)$ on the interval $[0,\pi ]$." The justification requires stating that $e^x$ and $\cos (x)$ are both continuous on $(-\infty ,\infty )$, and therefore their difference is also continuous on $(-\infty ,\infty )$, which includes the closed interval $[0,\pi ]$.

• Finding Constants in Piecewise Functions: "Find the value of $k$ that makes $f(x)=\{\begin{array}{ll}k{x}^2& x\le 2\\ 2x+k& x>2\end{array}$ continuous for all $x$." This requires ensuring continuity at the point $x=2$. You set the limit from the left equal to the limit from the right: $k(2{)}^2=2(2)+k⟹4k=4+k⟹3k=4⟹k=4/3$.

Common Mistakes

• Incorrectly Solving Inequalities: A very common algebra mistake when finding the domain for roots or logs. For $9-x^2\ge 0$, students might incorrectly conclude $x\ge 3$ or $x\ge \pm 3$ instead of the correct $-3\le x\le 3$.

• Mixing up $\ge$ and $>$: Forgetting that the argument of a logarithm must be strictly greater than zero ($>0$), while the radicand of a square root can be greater than or equal to zero ($\ge 0$).

• Ignoring Denominators Inside Other Functions: For a function like $f(x)=\sqrt{\frac{x}{x-2}}$, students might only check where $\frac{x}{x-2}\ge 0$ and forget that the inner denominator itself cannot be zero, so $x\ne 2$.

• Stating a Function is Continuous Without Justification: On a free-response question, you cannot simply state "the function is continuous." You must provide a reason based on its composition, such as "The function is a sum of a polynomial and a cosine function, both of which are continuous for all real numbers, so the function is continuous for all real numbers."