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: $f\left(5\right)+f\left(6\right)$

Answer 1

Given, $f\left(x\right)=\{\begin{array}{cc}4x^2-1;& -3\le x<2\\ 3x-2;& 2\le x\le 4\\ 2x-3;& 4<x\le 6\end{array}$

Then, $f\left(5\right)=2x-3=2\left(5\right)-3$
$=10-3=7$
and $f\left(6\right)=2\left(6\right)-3=12-3=9$
Therefore, $f\left(5\right)+f\left(6\right)=7+9=16$