Question

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

$\begin{array}{l}f\left(x\right)=\left\{\begin{array}{cc}{x}^2& \text{if\hspace{0.33em}}x<0\\ x+1& \text{if\hspace{0.33em}}x\ge 0\end{array}\right.\\ f\left(-2\right),f\left(-1\right),f\left(0\right),f\left(1\right),f\left(2\right)\end{array}$

Short answer

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

Step-by-step solution

Step 1. Given Information.

The given piecewise function is:

$f\left(x\right)=\left\{\begin{array}{cc}{x}^2& \text{if\hspace{0.33em}}x<0\\ x+1& \text{if\hspace{0.33em}}x\ge 0\end{array}\right.$

Here,$f\left(x\right)=x^2$ when$x<0$ and$f\left(x\right)=x+1$ when $x\ge 0$.

Step 2. Solve for first value.

Since $-2<0$, therefore substitute$x=-2$ in $f\left(x\right)=x^2$.

role="math" localid="1644519349680" $\begin{array}{c}f(-2)={\left(-2\right)}^2\\ =4\end{array}$

Step 3. Solve for second value.

Since $-1<0$, therefore substitute$x=-1$ in $f\left(x\right)=x^2$.

$\begin{array}{c}f(-1)={\left(-1\right)}^2\\ =1\end{array}$

Step 4. Solve for third value.

Since $0\ge 0$, therefore substitute$x=0$ in $f\left(x\right)=x+1$.

$\begin{array}{c}f\left(0\right)=0+1\\ =1\end{array}$

Step 5. Solve for fourth value.

Since $1\ge 0$, therefore substitute$x=1$ in $f\left(x\right)=x+1$.

$\begin{array}{c}f\left(1\right)=1+1\\ =2\end{array}$

Step 6. Solve for fifth value.

Since $2\ge 0$, therefore substitute$x=2$ in $f\left(x\right)=x+1$.

$\begin{array}{c}f\left(2\right)=2+1\\ =3\end{array}$

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