Problem 68 - 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 \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow x_{0}} f^{\prime}(x) .\) $$ \begin{array}{l}{\text { Let }} \\ {\qquad f(x)=\left\\{\begin{array}{ll}{x^{3}+\frac{1}{16},} & {x<\frac{1}{2}} \\\ {\frac{3}{4} x^{2},} & {x \geq \frac{1}{2}}\end{array}\right.} \\ {\text { Determine whether } f \text { is differentiable at } x=\frac{1}{2} . \text { If so, find }} \\ {\text { the value of the derivative there. }}\end{array} $$

Step-by-Step Solution

Answer

The function is differentiable at \( x = \frac{1}{2} \) with derivative \( \frac{3}{4} \).

Step 1: Check Continuity at x = 1/2

To verify the continuity of the function \( f \) at \( x = \frac{1}{2} \), evaluate the left-hand and right-hand limits of \( f(x) \) as \( x \) approaches \( \frac{1}{2} \). 1. Evaluate \( \lim_{x \to \frac{1}{2}^-} x^3 + \frac{1}{16} \): \[ \lim_{x \to \frac{1}{2}^-} \left(x^3 + \frac{1}{16}\right) = \left(\frac{1}{2}\right)^3 + \frac{1}{16} = \frac{1}{8} + \frac{1}{16} = \frac{1}{16} + \frac{2}{16} = \frac{3}{16} \]2. Evaluate \( \lim_{x \to \frac{1}{2}^+} \frac{3}{4} x^2 \): \[ \lim_{x \to \frac{1}{2}^+} \frac{3}{4} x^2 = \frac{3}{4} \left(\frac{1}{2}\right)^2 = \frac{3}{4} \cdot \frac{1}{4} = \frac{3}{16} \]Since both limits equal \( \frac{3}{16} \), \( f \) is continuous at \( x = \frac{1}{2} \).

Step 2: Find the Derivative of Each Piece

Determine the derivative of each piece of the piecewise function.1. For \( x < \frac{1}{2} \), the function is \( f(x) = x^3 + \frac{1}{16} \): \[ f'(x) = 3x^2 \]2. For \( x \geq \frac{1}{2} \), the function is \( f(x) = \frac{3}{4} x^2 \): \[ f'(x) = \frac{3}{2} x \]

Step 3: Evaluate the Limit of the Derivatives as x Approaches 1/2

Check the limit of the derivatives from both sides as \( x \to \frac{1}{2} \):1. Evaluate \( \lim_{x \to \frac{1}{2}^-} 3x^2 \): \[ \lim_{x \to \frac{1}{2}^-} 3x^2 = 3 \left(\frac{1}{2}\right)^2 = 3 \times \frac{1}{4} = \frac{3}{4} \]2. Evaluate \( \lim_{x \to \frac{1}{2}^+} \frac{3}{2} x \): \[ \lim_{x \to \frac{1}{2}^+} \frac{3}{2} x = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} \]Both limits equal \( \frac{3}{4} \), so \( \lim_{x \to \frac{1}{2}} f'(x) \) exists.

Step 4: Conclude Differentiability and Find the Derivative

Since \( f \) is continuous at \( x = \frac{1}{2} \) and \( \lim_{x \to \frac{1}{2}} f'(x) \) exists and is equal to \( \frac{3}{4} \), we can conclude by the given theorem that \( f \) is differentiable at \( x = \frac{1}{2} \).Thus, the value of the derivative at \( x = \frac{1}{2} \) is \( f'(\frac{1}{2}) = \lim_{x \to \frac{1}{2}} f'(x) = \frac{3}{4} \).

Key Concepts

Differentiability

Differentiability refers to whether a function has a derivative at a particular point. For a function to be differentiable at a point, it must be smooth and have no breaks, sharp corners, or jumps at that point. In other words, the function should behave in a predictable manner as you zoom in on the point.

When assessing differentiability for a piecewise function, it is crucial to check the function's behavior from either side of the point. If a function is differentiable at a point, the left-hand and right-hand derivatives both exist and are equal.

In the provided exercise, you are determining if the function is differentiable at a specific point, using the condition that the left and right derivatives at that point should be equal. This ensures that the function has a smooth transition at the point in question.

Continuity

Continuity in a function means that the function's graph is unbroken. You can trace the function from one point to another without lifting your pencil.

a function is continuous at a point if the following conditions are met:

• The function is defined at the point.
• The limit of the function as it approaches the point from the left equals the limit as it approaches from the right.
• The limit equals the function's value at that point.

In piecewise functions, checking the continuity at the boundaries involves calculating the one-sided limits as you approach from each piece of the function. In the original exercise, the function is determined to be continuous at the point of interest because the left and right limits match the function's single value at that exact point.

Derivative

A derivative is essentially a rate of change or a measure of how a function changes as its input changes. When you calculate a derivative at a point, you're finding the slope of the tangent line to the curve at that point. For piecewise functions, you must determine the derivative of each part separately.

For example, the exercise calculates the derivative of each piece of the function independently. Then, it checks whether the derivatives from the left and right sides at the point of interest match. If they do, that indicates the slope of the function is consistent across the boundary, thus helping in establishing differentiability.

Mean-Value Theorem

The Mean-Value Theorem (MVT) is a fundamental concept in calculus. It states that if a function is continuous over a closed interval and differentiable over an open interval, there exists at least one point within that interval where the derivative is equal to the overall average rate of change over the interval.

This theorem provides insight into the behavior of differentiable functions and tells us that somewhere between two points, the tangent of the curve will be parallel to the secant line joining the points. Although the MVT itself doesn't directly solve the exercise, it underpins the theorem used in the exercise suggesting that if the limit of the derivative exists at a point where continuity is already established, then differentiability can be concluded. This offers a theoretical foundation for understanding why the left and right derivatives must match for differentiability at a point.