Problem 68 - Calculus - AP Edition

Question

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0}\). You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section 4.8 ). Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0}\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow \infty} f^{\prime}(x)\) Let $$ f(x)=\left\\{\begin{array}{ll} x^{2}-16 x, & x<9 \\ \sqrt{x}, & x \geq 9 \end{array}\right. $$ Is \(f\) continuous at \(x=9 ?\) Determine whether \(f\) is differentiable at \(x=9 .\) If so, find the value of the derivative there.

Step-by-Step Solution

Answer

No, the function is not differentiable at x=9 since it is not continuous there.

Step 1: Check Continuity at x=9

First, ensure that the function is continuous at the point where the formulas change. For a function to be continuous at a point, \( \lim_{{x \to 9^-}} f(x) \) must equal \( \lim_{{x \to 9^+}} f(x) \) and \( f(9) \).Calculate: \( \lim_{{x \to 9^-}} f(x) = \lim_{{x \to 9^-}} (x^2 - 16x) = 9^2 - 16 \times 9 = -63 \)The value of \( f(9) = \sqrt{9} = 3 \).Calculate: \( \lim_{{x \to 9^+}} f(x) = \lim_{{x \to 9^+}} \sqrt{x} = \sqrt{9} = 3 \)Since \( \lim_{{x \to 9^-}} f(x) eq f(9) = \lim_{{x \to 9^+}} f(x) \, \) \( f \) is not continuous at \( x=9 \).

Step 2: Conclusion on Differentiability

Differentiability requires continuity at the point in question. Since we found in Step 1 that \( f(x) \) is not continuous at \( x=9 \, \) \( f(x) \) is not differentiable at \( x=9 \). There is no need to compute the derivative, as the continuity condition fails.

Key Concepts

Piecewise-Defined Functions

A piecewise-defined function is a function that is defined by different expressions depending on the value of the input variable. In simpler terms, it is like having different rules for different parts of the number line. This can make these functions a bit tricky when trying to analyze specific properties like continuity and differentiability.

For the given exercise, the function is defined differently for inputs less than 9 and for inputs greater than or equal to 9. Specifically, when \( x < 9 \), the expression \( x^2 - 16x \) is used, and for \( x \geq 9 \), the expression \( \sqrt{x} \) is used.

To analyze such a function, it is crucial to inspect the behavior at the boundary points where the rules change, in this case at \( x = 9 \). This point is the key juncture to check for passing over smoothly from one rule to the next, addressing whether the function is continuous and differentiable.

Continuity

Continuity at a point means that the function's value matches the values approached by the function from both sides of that point. This requires three conditions:

• The limit from the left side of the point exists.
• The limit from the right side of the point exists.
• These two limits, as well as the actual value of the function at the point, are the same.

In the case of our exercise at \( x = 9 \), the calculations showed that the limit approaching from the left is \( -63 \), and from the right is \( 3 \). The function's value at \( x = 9 \) is \( 3 \). However, because the left-hand limit does not equal the value of the function or the right-hand limit, the function \( f \) is not continuous at \( x = 9 \).

Continuity is a necessary condition for differentiability. If a function is not continuous at a certain point, it cannot be differentiable there either.

Limits

The concept of limits is essential when working with piecewise functions and examining their properties. A limit describes the value that a function approaches as the input approaches a certain point. It allows us to predict the behavior of the function around that point, even if the function is not defined precisely at the point.

In our exercise, we need to evaluate the limits as \( x \) approaches 9 from both sides:

• From the left, the expression \( x^2 - 16x \) gives a limit of \(-63\).
• From the right, the expression \( \sqrt{x} \) gives a limit of \(3\).

Since these limits are not equal, it indicates a discontinuity at that point. Hence, understanding limits helps identify whether a piecewise-defined function smoothly transitions from one formula to another and whether it is continuous at points of interest.