Problem 42 - Calculus Early Transcendentals

Question

Suppose \(f^{\prime}(x)<00>f^{\prime \prime}(x),\) for \(x>a .\) Prove that \(f\) is not differentiable at \(a\). (Hint: Assume \(f\) is differentiable at \(a\) and apply the Mean Value Theorem to \(f^{\prime} .\) ) More generally, show that if \(f^{\prime}\) and \(f^{\prime \prime}\) change sign at the same point, then \(f\) is not differentiable at that point.

Step-by-Step Solution

Answer

Short Answer: No, if \(f'\) and \(f''\) change sign at the same point \(a\), then \(f\) is not differentiable at that point.

Step 1: Assume \(f\) is differentiable at \(a\)

Let us assume that \(f\) is differentiable at \(a\) for the sake of contradiction. That means \(f'\) exists at \(a\).

Step 2: Apply the Mean Value Theorem on the intervals \((x, a)\) and \((a, x)\)

Since \(f\) is differentiable at \(a\), we can apply the Mean Value Theorem on the intervals \((x, a)\) for \(x < a\) and \((a, x)\) for \(x > a\). According to MVT, there must exist points \(c_1\) in \((x, a)\) and \(c_2\) in \((a, x)\) such that: 1. For \(x < a: f'(c_1) = \frac{f(x)-f(a)}{x-a}\) 2. For \(x > a: f'(c_2) = \frac{f(x)-f(a)}{x-a}\)

Step 3: Analyze the signs of the derivatives

From the given information, we know that: 1. For \(x < a\), we have \(f'(x) < 0\) and \(f''(x) > 0\). 2. For \(x > a\), we have \(f'(x) > 0\) and \(f''(x) < 0\). Now consider the consequences of these inequalities on \(f'(c_1)\) and \(f'(c_2)\): 1. Since \(f'(c_1) < 0\) and \(c_1 < a\), we must have \(\frac{f(x)-f(a)}{x-a} < 0\). This means \(f(x) < f(a)\). 2. Similarly, since \(f'(c_2) > 0\) and \(c_2 > a\), we must have \(\frac{f(x)-f(a)}{x-a} > 0\). This means \(f(x) > f(a)\).

Step 4: Show the contradiction

We found that for \(x < a\), \(f(x) < f(a)\), and for \(x > a\), \(f(x) > f(a)\). This implies that \(f\) achieves a local minimum and a local maximum at \(x=a\) simultaneously, which is clearly not possible. Therefore, we have reached a contradiction, and our original assumption that \(f\) is differentiable at \(a\) is false. This means that if \(f'\) and \(f''\) change sign at the same point, then \(f\) is not differentiable at that point.