Question

$\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

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

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 \(f(x)\).

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 \(x > 2\).

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 \(x < -2\).

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 \(1 < x < 2\).

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 \(f(x)\):

Combine all cases

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