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Chapter 2: Problem 3

$\begin{aligned} f(x) &= \begin{cases}\left|x^{2}-1\right|-1, & x \leq 1 \\\ |2 x-3|-|x-2|, & x>1\end{cases} \\ &=\left\\{\begin{array}{cc}x^{2}-2 & x \leq-1 \\ -x^{2} & -1

Short Answer

Expert verified
Question: What is the derivative of the given piecewise function $f(x) = \left\{ \begin{array}{cc} x^2-2 & x \leq -1 \\ -x^2 & -1 < x \leq 1 \\ 1-x & 1 < x \leq \frac{3}{2} \\ 3x-5 & \frac{3}{2} < x < 2 \\ x-1 & x \geq 2 \end{array} \right.$ Answer: The derivative of the given piecewise function is: $f'(x) = \left\{ \begin{array}{cc} 2x & x \leq -1 \\ -2x & -1 < x \leq 1 \\ -1 & 1 < x \leq \frac{3}{2} \\ 3 & \frac{3}{2} < x < 2 \\ 1 & x \geq 2 \end{array} \right.$

Step by step solution

01

Identify the regions and corresponding expressions for f(x).

We are given a piecewise function f(x) with different expressions for different intervals. Let's first specify the regions and their corresponding expressions: 1. Region 1: \(x \leq -1\) with expression \(f(x) = x^2-2\) 2. Region 2: \(-1 < x \leq 1\) with expression \(f(x) = -x^2\) 3. Region 3: \(1 < x \leq \frac{3}{2}\) with expression \(f(x) = 1-x\) 4. Region 4: \(\frac{3}{2} < x < 2\) with expression \(f(x) = 3x-5\) 5. Region 5: \(x \geq 2\) with expression \(f(x) = x-1\)
02

Calculate the derivatives for each expression.

Now, we will find the derivative for each expression with respect to x: 1. Region 1: \(f'(x) = \frac{d}{dx}(x^2-2) = 2x\) 2. Region 2: \(f'(x) = \frac{d}{dx}(-x^2) = -2x\) 3. Region 3: \(f'(x) = \frac{d}{dx}(1-x) = -1\) 4. Region 4: \(f'(x) = \frac{d}{dx}(3x-5) = 3\) 5. Region 5: \(f'(x) = \frac{d}{dx}(x-1) = 1\)
03

Combine the derivatives to generate the final result.

The final step is to put everything together and write down the piecewise function for the derivative, f'(x), using the results obtained in step 2: $f'(x) = \left\{ \begin{array}{cc} 2x & x \leq -1 \\ -2x & -1 < x \leq 1 \\ -1 & 1 < x \leq \frac{3}{2} \\ 3 & \frac{3}{2} < x < 2 \\ 1 & x \geq 2 \end{array} \right.$ So, the derivative of the given piecewise function is: $f'(x) = \left\{ \begin{array}{cc} 2x & x \leq -1 \\ -2x & -1 < x \leq 1 \\ -1 & 1 < x \leq \frac{3}{2} \\ 3 & \frac{3}{2} < x < 2 \\ 1 & x \geq 2 \end{array} \right.$

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Most popular questions from this chapter

$\mathrm{f}(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{(2 \sin \mathrm{x})^{2 \mathrm{n}}}{3^{\mathrm{n}}-(2 \cos \mathrm{x})^{2 \mathrm{n}}}\( \)\Rightarrow \lim _{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{1-\left(\frac{2}{\sqrt{3}} \cos x\right)^{2 n}}$ \(\mathrm{f}(\mathrm{x})\) is discontinuous whenever \(2 \sin x=\pm 1 \& \frac{2}{\sqrt{3}} \cos x=\pm 1\) \(\Rightarrow \mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{6}\)

A) $f(x)= \begin{cases}\frac{1+a \cos x}{x^{2}}, & x<0 \\ b \tan \left(\frac{\pi}{[x+3]}\right), & x \geq 0\end{cases}$ \(\lim _{x \rightarrow 0^{-}} \frac{1+a \cos x}{x^{2}}\) Exists if \(\mathrm{a}=-1\). Then limit is \(\frac{1}{2}\). $\lim _{x \rightarrow 0^{-}} b+a \frac{\pi}{3}=\frac{1}{2} \Rightarrow b=\frac{1}{2 \sqrt{3}} \Rightarrow[a-2 b]=-2$ Hence, \(\mathrm{R}\) is correct. B) $\lim _{x \rightarrow-\frac{\pi}{2}}(-2 \sin x)=2$ \(\lim _{x \rightarrow-\frac{\pi^{*}}{2}}(a \sin x+b)=+b-a\) \(\lim _{x \rightarrow \frac{\pi}{2}}(\operatorname{asin} x+b)=a+b\) \(\lim _{x \rightarrow \frac{\pi}{2}} \cos x=0\) Hence \(\mathrm{P}, \mathrm{Q}\) is correct. C) $\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{3}{2}\right)^{\frac{\cos 3 x}{\cot 2 x}}=1$ $\lim _{x \rightarrow \frac{\pi^{-}}{2}}(1+|\cos x|)^{\frac{a|t a n x|}{b}}=e^{\frac{\lim _{T}}{x} \frac{a \sin x \mid}{b}}=e^{\frac{a}{b}}$ \(f\left(\frac{\pi}{2}\right)=\frac{\pi}{2}+3\) \(\Rightarrow \quad a=0\) D) \(\lim _{x \rightarrow 0^{0}} a+\frac{\sin [x]}{x}=a\). \(\lim _{x \rightarrow 0} b+\left[\frac{\sin x-x}{x^{3}}\right]=b-1\) \(f(0)=2\) \(\Rightarrow \mathrm{a}=2, \quad \mathrm{~b}=3\) So, \(\mathrm{T}\) is correct.

\(f(x)=\lim _{x \rightarrow \infty} x \tan ^{-1} n x\) $\left\\{\begin{array}{cl}\frac{\pi}{2} x, & x>0 \\ 0, & x=0 \\\ \frac{-\pi}{2} x, & x<0\end{array}\right.$

$\lim _{x \rightarrow a}(\cos (x-a))^{\frac{6}{(x-a) 2 \sin (x-a) \cos (x-a)}}$ $\Rightarrow e^{-\lim 2 \sin ^{2} \frac{(x-a)}{2} \times \frac{6}{(x-a) \sin (x-a) \cos (x-a)}}$ \(\Rightarrow e^{-3 / 2}\) Now, $h(x)=\lim _{n \rightarrow \infty} \frac{\sin \frac{x}{2^{n}}}{\frac{x}{2^{n}}} \times x=x$ \(h(a)=a\) \(\Rightarrow \mathrm{k}=\frac{3}{2}\) Hence, \(\mathrm{C}\) is correct.

$\lim _{x \rightarrow 0^{\prime}} \cos \left(x \cos \frac{1}{x}\right)=1 \quad f(0)$ is not defined. \(\lim _{x \rightarrow 0^{-}} \cos \left(x \cos \frac{1}{x}\right)=1\)
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