Chapter 2.1, Problem 27E

To determine

To evaluate $f\left(-2\right),f\left(-1\right),f\left(0\right),f\left(1\right),f\left(2\right)$ valuesby using the piecewise function $f\left(x\right)=\{\begin{array}{l}{x}^2,\text{ if }x<0\\ x+1,\text{ if }x\ge 0\end{array}$

Answer to Problem 27E

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

Explanation of Solution

Given information:consider the piecewise function $f\left(x\right)=\{\begin{array}{l}{x}^2,\text{ if }x<0\\ x+1,\text{ if }x\ge 0\end{array}$

Calculation:

First find the value of $f\left(-2\right),f\left(-1\right)$

Since the valuesof $x=-2<0$ and $x=-1<0$

From the above piecewise function,to be taken as

  $f\left(x\right)=x^2$

  $\begin{array}{l}f\left(-2\right)={\left(-2\right)}^2\\ \Rightarrow f\left(-2\right)=4\end{array}$

And,

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

Find the values of $f\left(0\right),f\left(1\right),f\left(2\right)$

Since the values of $x=0,x=1>0$ and $x=2>0$

  $f\left(x\right)=x+1$

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

And,

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

Also,

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

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