Question

A function $f:[-3,7)\to R$ is defined as follows
$f\left(x\right)=\{\begin{array}{cc}4x^2-1;& -3\le x<2\\ 3x-2;& 2\le x\le 4\\ 2x-3;& 4<x<7\end{array}$
(i) $f\left(5\right)+f\left(6\right)$
(ii) $f\left(1\right)-f\left(3\right)$
(iii) $f(-2)-f(-4)$
(iv) $\frac{f\left(3\right)+f\left(1\right)}{2f\left(6\right)-f\left(1\right)}$

Answer 1

$\left(i\right)$ $f\left(5\right)=2\left(5\right)-3=7f\left(6\right)=2\left(6\right)-3=9\therefore f\left(5\right)+f\left(6\right)=7+9=16$

$\left(ii\right)$ $f\left(1\right)=4\left(1^2\right)-1=3f\left(3\right)=3\left(3\right)-2=7\therefore f\left(1\right)-f\left(3\right)=3-7=-4$

$\left(iii\right)$ $f(-4)$ is not defined because $-4$is not in domain.

$\left(iv\right)$ $f\left(3\right)=3\left(3\right)-2=7f\left(1\right)=4\left(1^2\right)-1=3f\left(6\right)=2\left(6\right)-3=9\therefore \frac{f\left(3\right)+f\left(1\right)}{2f\left(6\right)-f\left(1\right)}=\frac{7+3}{18-3}=\frac{2}{3}$