Problem 66 - Calculus Early Transcendentals

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 \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow x_{0}} f^{\prime}(x) .\) $$ \begin{array}{l}{\text { Let } \quad f(x)=\left\\{\begin{array}{ll}{x^{2}-16 x,} & {x<9} \\ {\sqrt{x},} & {x \geq 9}\end{array}\right.} \\ {\text { Is } f \text { continuous at } x=9 \text { ? Determine whether } f \text { is differ- }} \\ {\text { entiable at } x=9 . \text { If so, find the value of the derivative }} \\ {\text { there. }}\end{array} $$

Step-by-Step Solution

Answer

The function is not differentiable at \( x=9 \) because it is not continuous there.

Step 1: Checking Continuity at x=9

To verify the continuity of the function at \( x=9 \), evaluate the left-hand limit and the right-hand limit, and compare them with the value of the function at \( x=9 \). For the left-hand limit when \( x<9 \), we use the formula \( f(x) = x^2 - 16x \). As \( x \to 9^- \), \[ \lim_{x \to 9^-} x^2 - 16x = 9^2 - 16 \cdot 9 = 81 - 144 = -63. \]For the right-hand limit when \( x \ge 9 \), we use the formula \( f(x) = \sqrt{x} \). As \( x \to 9^+ \), \[ \lim_{x \to 9^+} \sqrt{x} = \sqrt{9} = 3. \]Since \( f(9) = \sqrt{9} = 3 \), the function is not continuous at \( x=9 \) because the left-hand limit \(-63\) and the right-hand limit \(3\) are not equal.

Step 2: Conclusion about Differentiability

Given that \( f(x) \) is not continuous at \( x=9 \), the function cannot be differentiable at this point. Differentiability requires the function to be continuous, so no further checks for the derivative are needed.

Key Concepts

Piecewise-Defined Functions

Piecewise-defined functions are mathematical functions defined by different expressions or formulas over different parts of their domain. This means that a single function can "switch" formulas depending on the value of the input variable. Such functions are common in modeling real-world situations where a behavior changes at a certain point, like tax brackets or temperature scales.

When analyzing piecewise-defined functions, it's important to:

• Identify the sub-domains: These are the intervals where different formulas are applied.
• Know the "breakpoint": This is where the formula changes. For instance, in the function described, the breakpoint is at 9.
• Evaluate at the boundaries: Check how the function transitions from one formula to another at the breakpoint.

Understanding piecewise-defined functions is crucial, particularly when checking for continuity and differentiability, as different segments must be analyzed independently and then collectively at the breakpoint.

Continuity at a Point

A function is continuous at a particular point if it has no sudden jumps, breaks, or holes at that point. Continuity of a piecewise-defined function at a certain point involves evaluating limits from both sides of the point and ensuring they, together with the function's value at the point, are equal. This requires that:

• The left-hand limit as the input approaches the point from the left should equal the right-hand limit as the input approaches from the right.
• The value of the function at that point should be equal to these limits.

In the exercise, to check continuity at the breakpoint 9, we calculated both the left-hand limit, using the formula for when the input is less than 9, and the right-hand limit, using the formula for inputs equal to or greater than 9. Since these limits did not match, the function was not continuous at 9. For a function to be differentiable at a point, it must also be continuous there; hence, discontinuity implies non-differentiability.

Mean-Value Theorem

The Mean-Value Theorem (MVT) is a fundamental theorem in calculus that links the concepts of continuity and differentiability. Essentially, it states that if a function is continuous on a closed interval and differentiable on the open interval, there exists at least one point within the interval where the instantaneous rate of change (derivative) equals the average rate of change over the entire interval.

For piecewise-defined functions, using the MVT can be insightful, especially when determining if a function is differentiable at a boundary between two different pieces. The theorem assumes that within the interval, the function is smooth and does not have any sharp turns. If discontinuity is found at the boundary, as in our exercise, it invalidates differentiability according to the MVT because the requirement for continuity at the point is not met.

The step-by-step solution highlights that when continuity fails at a point, checking for differentiability using the MVT becomes unnecessary since discontinuity already implies non-differentiability.