Chapter 2.4, Problem 26AYU

If $f\left(x\right)=\{\begin{array}{l}-3x\text{if}x<-1\\ 0\text{if}x=-1\\ 2x^2+1\text{if}x>-1\end{array}$

find: (a) $f\left(-2\right)$

(b) $f\left(-1\right)$

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

To determine

To find: The values of $f\left(-2\right),f\left(-1\right)$ and $f\left(0\right)$ using the given piecewise defined function $f\left(x\right)=\{\begin{array}{c}-3x if x<-1\\ 0 ifx=-1\\ 2x^2+1 ifx>-1\end{array}$

Answer to Problem 26AYU

Solution:

The values of $f\left(-2\right),f\left(-1\right)$ and $f\left(0\right)$ using the given piecewise defined function are 6, 0 and 1.

Explanation of Solution

Given:

The function is: $f\left(x\right)=\{\begin{array}{c}-3x if x<-1\\ 0 ifx=-1\\ 2x^2+1 ifx>-1\end{array}$

It is asked to find the values of $f\left(-2\right),f\left(-1\right)$ and $f\left(0\right)$ using the piecewise defined function.

1. Consider $f\left(-2\right),-2$ belongs to the condition $x<-1$.

$f\left(x\right)=-3x$

$f\left(-2\right)=-3\left(-2\right)=6$

2. Consider $f\left(-1\right),-1$ belongs to the condition $x=-1$.

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

$f\left(-1\right)=0$

3. Consider $f\left(0\right),0$ belongs to the condition $x>-1$.

$f\left(x\right)=2x^2+1$

$f\left(0\right)=2{\left(0\right)}^2+1=1$

The values of $f\left(-2\right),f\left(-1\right)$ and $f\left(0\right)$ using the given piecewise defined function are 6, 0 and 1.