Question

A function $f:(-3,7)\to R$ is defined as follows:
$f\left(x\right)=\{\begin{array}{cc}4x^2-1;& -3\le x<2\\ 3x-2;& 2\le x\le 4\\ 2x-3;& 4<x\le 6\end{array}$
Find:
$f\left(1\right)-f(-3)$

Answer 1

Given, $f:(-3,7)\to R$ is defined as follows:
$f\left(x\right)=\{\begin{array}{cc}4x^2-1;& -3\le x<2\\ 3x-2;& 2\le x\le 4\\ 2x-3;& 4<x\le 6\end{array}$

Then, $f\left(1\right)=4(1{)}^2-1=4(1)-1=4-1=3$
and $f(-3)=4(-3{)}^2-1=4(9)-1=36-1=35$
Therefore, $f\left(1\right)-f(-3)=3-35=-32$