Question

Piecewise Defined Functions Evaluate the piecewise defined function at the indicated values.

$\begin{array}{l}f\left(x\right)=\left\{\begin{array}{l}{x}^2+2x\mathrm{if}x\le -1\\ x\mathrm{if}-1<x\le 1\\ -1\mathrm{if}x>1\end{array}\right.\\ f\left(-4\right),f\left(-\frac{3}{2}\right),f\left(-1\right),f\left(0\right),f\left(25\right)\end{array}$

Short answer

The required values are $f\left(-4\right)=8,f\left(-\frac{3}{2}\right)=-\frac{3}{4},f\left(-1\right)=-1,f\left(0\right)=0,f\left(25\right)=-1$.

Step-by-step solution

Step 1. Given Information.

The given piecewise function is:

$f\left(x\right)=\left\{\begin{array}{l}{x}^2+2x\mathrm{if}x\le -1\\ x\mathrm{if}-1<x\le 1\\ -1\mathrm{if}x>1\end{array}\right.$

Here,$f\left(x\right)=x^2+2x$ when $x\le -1$,$f\left(x\right)=x$ when$-1<x\le 1$ and$f\left(x\right)=-1$ when $x>1$.

Step 2. Solve for first value.

Since $-4\le -1$, therefore substitute$x=-4$ in $f\left(x\right)=x^2+2x$.

width="156" height="71" role="math">f(−4)=−42+2−4=16−8=8

Step 3. Solve for second value.

Since $-\frac{3}{2}\le -1$, therefore substitute$x=-\frac{3}{2}$ in $f\left(x\right)=x^2+2x$.

$$\begin{gathered} f\left(-\frac{3}{2}\right)={\left(-\frac{3}{2}\right)}^2+2\left(-\frac{3}{2}\right) \\ =\frac{9}{4}-3 \\ =\frac{9-12}{4} \\ =-\frac{3}{4} \end{gathered}$$

Step 4. Solve for third value.

Since $-1\le -1$, therefore substitute$x=-1$ in $f\left(x\right)=x^2+2x$.

width="156" height="71" role="math">f(−1)=−12+2−1=1−2=−1

Step 5. Solve for fourth value.

Since $-1<0\le 1$, therefore substitute$x=0$ in $f\left(x\right)=x$.

$f\left(0\right)=0$

Step 6. Solve for fifth value.

Since $25>1$, therefore substitute$x=25$ in $f\left(x\right)=-1$.

$f\left(25\right)=-1$

Thus, the required values are $f\left(-4\right)=8,f\left(-\frac{3}{2}\right)=-\frac{3}{4},f\left(-1\right)=-1,f\left(0\right)=0,f\left(25\right)=-1$.