Problem 69 - 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 $$ f(x)=\left\\{\begin{array}{ll} x^{2}, & x \leq 1 \\ \sqrt{x}, & x>1 \end{array}\right. $$ Determine whether \(f\) is differentiable at \(x=1 .\) If so, find the value of the derivative there.

Step-by-Step Solution

Answer

The function \(f\) is not differentiable at \(x=1\).

Step 1: Check Continuity at x=1

For a function to be differentiable, it must first be continuous at that point. Calculate \(\lim_{x \to 1^-} f(x)\) and \(\lim_{x \to 1^+} f(x)\). The left-hand limit when \(x \leq 1\) is \(x^2\), so \(\lim_{x \to 1^-} f(x) = 1^2 = 1\). The right-hand limit when \(x > 1\) is \(\sqrt{x}\), so \(\lim_{x \to 1^+} f(x) = \sqrt{1} = 1\). Also, \(f(1) = 1^2 = 1\). Therefore, \(f\) is continuous at \(x=1\) since \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)\).

Step 2: Determine Derivative from Left Side

Calculate the derivative for \(x \leq 1\), which is \(f(x) = x^2\). The derivative is \(f'(x) = 2x\). At \(x = 1\), \(f^{\prime}(1^{-}) = 2(1) = 2\).

Step 3: Determine Derivative from Right Side

Now calculate the derivative for \(x > 1\), which is \(f(x) = \sqrt{x}\). The derivative is \(f'(x) = \frac{1}{2\sqrt{x}}\). As \(x\) approaches 1 from the right, \(f^{\prime}(1^{+}) = \frac{1}{2\sqrt{1}} = \frac{1}{2}\).

Step 4: Check Limits of Derivative from Both Sides

For differentiability at \(x=1\), \(\lim_{x \to 1^-} f^{\prime}(x)\) and \(\lim_{x \to 1^+} f^{\prime}(x)\) must be equal. From Steps 2 and 3, we have \(f^{\prime}(1^{-}) = 2\) and \(f^{\prime}(1^{+}) = \frac{1}{2}\). Since these are not equal, \(f\) is not differentiable at \(x=1\).

Key Concepts

Differentiability

Differentiability describes a function's smoothness at a certain point. If a function is differentiable at a point, its graph will not have any sharp corners or cusps at that location. To determine differentiability, two main criteria must be satisfied:

• The function must be continuous at the point being evaluated.
• The limits of the derivative from both sides of the point must be equal.

For piecewise-defined functions, such as in our exercise, it's crucial to evaluate these criteria specifically at the points where the defining rules of the function change. In our example, we checked differentiability at \( x = 1 \). Despite the function being continuous at this point, it was not differentiable because the left-hand derivative (from \( x \leq 1 \)) and right-hand derivative (from \( x > 1 \)) did not match, rendering the function non-differentiable at \( x = 1 \).

Continuity

Continuity of a function at a specific point is foundational for differentiability. A function is continuous at \( x = x_0 \) if there are no jumps, holes, or asymptotes at that point. This means that the limit of the function as it approaches \( x_0 \) from the left (\( \lim_{x o x_0^-} f(x)\)) is equal to the limit from the right (\( \lim_{x o x_0^+} f(x)\)), and both are equal to the function's value at that point \( f(x_0) \).

In the case of the given piecewise function, we verified that the left-hand limit and the right-hand limit at \( x = 1 \) both equaled 1, just like \( f(1) \). Hence, the function is continuous at \( x = 1 \). However, continuity alone does not ensure differentiability, as the derivatives from both sides of the point must also align.

Derivative

The derivative of a function gives us the rate of change or the slope of the function at any point. It is essentially the function's 'sensitivity' to change at a specific value of \(x\). For piecewise-defined functions, the derivative can differ for each section of the function's domain. It is crucial to calculate the derivative for each piece separately.

Let's consider our step-by-step solution:

• For the section where \( x \leq 1 \), our function \( f(x) = x^2 \) results in a derivative \( f'(x) = 2x \).
• For the section where \( x > 1 \), the function \( f(x) = \sqrt{x} \) yields a derivative \( f'(x) = \frac{1}{2\sqrt{x}} \).

At \( x = 1 \), these expressions provide different derivative values: \(2\) from the left and \(\frac{1}{2}\) from the right. As these derivative values differ, the function is not differentiable at that point, highlighting that both continuity and matching derivatives are essential for differentiability.