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Chapter 4: Problem 54

$\begin{aligned} &f(x)=\left|x^{2}-3\right| x|+2| \\ &f(x)= \begin{cases}x^{2}-3 x+2, & 02 \\ x^{2}+3 x+2, & -1

Short Answer

Expert verified
Question: Find the simplified expression for the function \(f(x) = |x^2 - 3x| + |x| + 2\) in each given interval. Answer: The simplified expression for the function \(f(x)\) is: $$ f(x)= \begin{cases} x^{2}-3 x+2, & 02 \\ x^{2}+3 x+2, & -1<x<0 \text { or } x<-2 \\ -\left(x^{2}-3 x+2\right), & 1<x<2 \\ -\left(x^{2}+3 x+2\right), & -2<x<-1 \end{cases} $$

Step by step solution

01

Define the cases for the function

To start, let's define the four different cases based on the exercise's given intervals: 1. \(0 < x < 1\) or \(x > 2\) 2. \(-1 < x < 0\) or \(x < -2\) 3. \(1 < x < 2\) 4. \(-2 < x < -1\) For each of these cases, we will find the corresponding formula for the function \(f(x)\).
02

Solve for Case 1

In this interval, \(x^2 - 3x\) is non-negative, so the absolute value of it won't change its sign. Therefore, \(f(x) = x^2 - 3x + 2\) for \(0 < x < 1\) or \(x > 2\).
03

Solve for Case 2

In this interval, \(x^2 - 3x\) is negative, so the absolute value will change its sign. However, \(x\) is also negative in this interval, so the sign of \(x\) will also change. Therefore, we have \(f(x) = -\left(-(x^2 - 3x)\right) + 2 = x^2 + 3x + 2\) for \(-1 < x < 0\) or \(x < -2\).
04

Solve for Case 3

In this interval, \(x^2 - 3x\) is positive, so the absolute value won't change its sign. However, since \(x\) is also positive in this interval, its sign will change. Therefore, we have \(f(x) = -\left(x^2 - 3x\right) + 2\) for \(1 < x < 2\).
05

Solve for Case 4

In this interval, \(x^2 - 3x\) is negative, so the sign of \(x^2 - 3x\) will change. \(x\) is also negative in this interval, so the sign of \(x\) will change as well. Therefore, we have \(f(x) = -\left(-(x^2 - 3x)\right) + 2 = x^2 + 3x + 2\) for \(-2 < x < -1\). Now, combining all the cases, we have the final expression for \(f(x)\):
06

Combine all cases

The final expression for \(f(x)\), based on the different cases, is: $$ f(x)= \begin{cases} x^{2}-3 x+2, & 02 \\ x^{2}+3 x+2, & -1<x<0 \text { or } x<-2 \\ -\left(x^{2}-3 x+2\right), & 1<x<2 \\ -\left(x^{2}+3 x+2\right), & -2<x<-1 \end{cases} $$

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Most popular questions from this chapter

\(\sqrt{\mathrm{y}+\mathrm{x}}+\sqrt{\mathrm{y}-\mathrm{x}}=\mathrm{c}\) Squaring both sides. \(2 y+2 \sqrt{y^{2}-x^{2}}=c^{2}\) Differentiating $2 y^{\prime}+\frac{2\left(y y^{\prime}-x\right)}{\sqrt{y^{2}-x^{2}}}=0$ \(y^{\prime}=\frac{x}{y+\sqrt{y^{2}-x^{2}}}=\frac{2 x}{c^{2}}\) Rationilising, we get \(y^{\prime}=\frac{y-\sqrt{y^{2}-x^{2}}}{x}\)

\(f(x)=\frac{x}{1+e^{1 / x}}\) \(f(x)\left(1+e^{1 / x}\right)-x=0\) \(f^{\prime}(x)\left(1+e^{1 / x}\right)-\frac{f(x) e^{l / x}}{x^{2}}-1=0\) \(x^{2} f^{\prime}(x) \frac{(x)}{f(x)}-f(x) e^{1 / x}-x^{2}=0\)

\(f(x)=e^{a x}+e^{b x}\) \(f^{\prime}(x)=a e^{a x}+b e^{b x}\) \(f^{\prime \prime}(x)=a^{2} e^{a x}+b^{2} e^{b x}\) Now \(f^{\prime \prime}(x)-2 f^{\prime}(x)-15 f(x)=0\) $\Rightarrow\left(a^{2}-2 a-15\right) e^{a x}+\left(b^{2}-2 b-15\right) e^{b x}=0$ \(\Rightarrow a^{2}-2 a-15=0 \quad \& \quad b^{2}-2 b-15=0\) a, bare two roots of \(x^{2}-2 x-15=0\) \(\mathrm{ab}=-15\)

\(x=a \cos t+\frac{b}{2} \cos 2 t, \quad y=a \sin t+\frac{b}{2} \sin 2 t\) $\frac{d x}{d t}=-a \sin t-b \sin 2 t, \quad \frac{d y}{d t}=a \cos t+b \cos 2 t$ \(\frac{d y}{d x}=\frac{-(a \cos t+b \cos 2 t)}{a \sin t+b \sin 2 t}\) $\frac{d^{2} y}{d x^{2}}=\left[\begin{array}{l}\frac{(a \sin t+b \sin 2 t)(+a \sin t+2 b \sin 2 t)}{(a \sin t+b \sin 2 t)^{2}} \\ +(a \cos t+b \cos 2 t)(a \cos t+2 b \cos 2 t)\end{array}\right] \times \frac{1}{-(a \sin t+b \sin 2 t)}$ For \(\frac{d^{2} y}{d x^{2}}=0\) \((a \sin t+b \sin 2 t)(a \sin t+2 b \sin 2 t)+(a \cos t+b \cos 2 t)\) \((a \cos t+2 b \cos 2 t)=0\) \(\Rightarrow a^{2}+2 b^{2}+a b \cos t+2 a b \cos t=0\) \(\Rightarrow \cos t=\frac{-\left(a^{2}+2 b^{2}\right)}{3 a b}\)

If \(\mathrm{f}(\mathrm{x})=|\ln | \mathrm{x} \|\), then \(\mathrm{f}^{\prime}(\mathrm{x})\) equals (A) \(\frac{-\operatorname{sgn} \mathrm{x}}{|\mathrm{x}|}\), for \(|\mathrm{x}|<1\), where \(\mathrm{x} \neq 0\) (B) \(\frac{1}{x}\) for \(|x|>1\) and \(-\frac{1}{x}\) for \(|x|<1, x \neq 0\) (C) \(-\frac{1}{x}\) for \(|x|>1\) and \(\frac{1}{x}\) for \(|x|<1\) (D) \(\frac{1}{x}\) for \(|x|>0\) and \(-\frac{1}{x}\) for \(x<0\)
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