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Chapter 3: Problem 7

The set of all points where \(f(x)=\sqrt[3]{x^{2}|x|}-|x|-1\) is not differentiable is (A) \(\\{0\\}\) (B) \(\\{-1,0,1\\}\) (C) \(\\{0,1\\}\) (D) none of these

Short Answer

Expert verified
Answer: The function is not differentiable at the set \(\{0\}\).

Step by step solution

01

Break down the function

We can break down \(f(x)=\sqrt[3]{x^{2}|x|}-|x|-1\) into separate functions: \(g(x)=\sqrt[3]{x^{2}|x|}\), \(h(x)=-|x|\), and \(k(x)=-1\). We can rewrite \(f(x)\) as a combination of these functions: \(f(x)=g(x)+h(x)+k(x)\).
02

Find the derivatives of the separate functions

Let's find the derivatives of the separate functions: (1) \(g(x)=\sqrt[3]{x^{2}|x|}\): $$g'(x)=\frac{d}{dx}\left(\sqrt[3]{x^{2}|x|}\right)=\frac{2x|x|+x^2\text{sgn}(x)}{3\sqrt[3]{(x^{2}|x|)^2}},$$ where \(\text{sgn}(x)\) is the signum function. Note that \(g'(x)\) is not defined at \(x=0\). (2) \(h(x)=-|x|\): $$h'(x)=\frac{d}{dx}\left(-|x|\right)=\begin{cases}1&,x<0\\-1&,x>0\\ \text{undefined}&,x=0\end{cases}.$$ (3) \(k(x)=-1\): The derivative of a constant function is 0, so \(k'(x)=0\) for all \(x\).
03

Combine the derivatives of g, h, and k

To find the derivative of \(f(x)=g(x)+h(x)+k(x)\), we can sum the derivatives of the separate functions we found in Step 2: $$f'(x)=g'(x)+h'(x)+k'(x)=\frac{2x|x|+x^2\text{sgn}(x)}{3\sqrt[3]{(x^{2}|x|)^2}}+h'(x).$$
04

Investigate the undefined points

Now, we must determine which points have undefined derivatives, which means the function is not differentiable at these points. For \(x \neq 0, f'(x)\) is well-defined since \(g'(x)\) is defined for \(x\neq 0\), \(h'(x)\) is constant, and \(k'(x)\) is constant. For \(x=0\), we have already observed that both \(g'(0)\) and \(h'(0)\) are undefined. Therefore, \(f'(0)\) is undefined as well.
05

Determine the set of points where f(x) is not differentiable

We have concluded that \(f'(x)\) is undefined only at \(x=0\). Thus, the set of all points where the function is not differentiable is \(\{0\}\). The correct answer is (A).

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Most popular questions from this chapter

If \(f(x)=\frac{x}{1+e^{1 / x}}, x \neq 0\) and \(f(0)=0\) then, (A) \(f(x)\) is continuous at \(x=0\) and \(f^{\prime}(x)=1\) (B) \(\mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=0\) (C) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\) and \(\mathrm{f}^{\prime}(\mathrm{x})\) does not exists (D) \(f(x)\) is continuous at \(x=0\) and \(f^{\prime}(x)=0\)

Consider the function \(f(x)\) $= \begin{cases}x^{3} & \text { if } x<0 \\\ x^{2} & \text { if } 0 \leq x<1 \\ 2 x-1 & \text { if } 1 \leq x<2 \\\ x^{2}-2 x+3 & \text { if } x \geq 2\end{cases}$ then \(\mathrm{f}\) is continuous and differentiable for (A) \(x \in R\) (B) \(x \in R-\\{0,2\\}\) (C) \(x \in \mathbb{R}-\\{2\\}\) (D) \(x \in \mathbb{R}-\\{1,2\\}\)

Let $f(x)=\left[\begin{array}{ll}\frac{3 x^{2}+2 x-1}{6 x^{2}-5 x+1} & \text { for } x \neq \frac{1}{3} \\ -4 & \text { for } x=\frac{1}{3}\end{array}\right.\( then \)f^{\prime}\left(\frac{1}{3}\right)$ (A) is equal to-9 (B) is equal to \(-27\) (C) is equal to 27 (D) does not exist

$f(x)=\left\\{\begin{array}{ll}\frac{x}{2 x^{2}+|x|} & , x \neq 0 \\ 1 & , x=0\end{array}\right.\( then \)f(x)$ is (A) Continuous but non-differentiable at \(\mathrm{x}=0\) (B) Differentiable at \(\mathrm{x}=0\) (C) Discontinuous at \(\mathrm{x}=0\) (D) None of these

If \(f(x)=\operatorname{Max} \cdot\\{1,(\cos x+\sin x)\) \((\sin x-\cos x)\\} 0 \leq x \leq 5 \pi / 4\), then (A) \(\mathrm{f}(\mathrm{x})\) is not differentiable at \(\mathrm{x}=\pi / 6\) (B) \(f(x)\) is not differentiable at \(x=5 \pi / 6\) (C) \(f(x)\) is continuous for \(x \in[0,5 \pi / 4]\) (D) None of these
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