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Chapter 4: Problem 40

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.

Short Answer

Expert verified
Question: Prove that if both the first and second derivatives of a function change their signs at the same point, the function is not differentiable at that point. Answer: To prove this result, we first assume that the function is differentiable at the given point and then apply the Mean Value Theorem to the first derivative for values less than and greater than the given point. By comparing the values of the first derivative, we find a contradiction, indicating that our initial assumption was incorrect. Therefore, if both the first and second derivatives change their signs at the same point, the function is not differentiable at that point.

Step by step solution

01

Assume the function is differentiable at a

Let's assume that \(f\) is differentiable at \(a\). Then, \(f'(a)\) exists.
02

Apply Mean Value Theorem to \(f'(x)\) for \(x

Since \(f\) is differentiable at \(a\), we can apply the Mean Value Theorem (MVT) to the first derivative \(f'(x)\) for \(x 0\) for \(x < a\). Therefore, \(f'(c_1) > 0\) and we have \(f'(a) > f'(x)\) for \(x < a\).
03

Apply Mean Value Theorem to \(f'(x)\) for \(x>a\)

Similarly, we can apply the MVT to the first derivative \(f'(x)\) for \(x>a\). There exists a point \(c_2\) in the interval \((a, x)\) such that $$ f'(c_2) = \frac{f'(x) - f'(a)}{x - a}. $$ By the given condition, \(f'(x) > 0\) and \(f''(x) < 0\) for \(x > a\). Therefore, \(f'(c_2) < 0\) and we have \(f'(a) < f'(x)\) for \(x > a\).
04

Show a contradiction

In steps 2 and 3, we have shown that for \(x f'(x)\) and for \(x>a\), we have \(f'(a) < f'(x)\). This means that \(f'(a)\) must be both greater than and less than all values of \(f'(x)\) for \(x \neq a\), which leads to a contradiction. Thus, our assumption that \(f\) is differentiable at \(a\) is false.
05

Generalization

Now, let's generalize the result. Suppose the first and second derivatives change signs at the same point \(p\). Then for some neighborhood of \(p\), we have \(f'(x)<00>f''(x)\) for \(x>p\). Proceeding similarly to steps 2, 3, and 4, we can show that a contradiction arises if we assume that \(f\) is differentiable at \(p\). Therefore, if both first and second derivatives change their signs at the same point, the function is not differentiable at that point.

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