Question

A function $f:[-7,6)\to R$ is defined as follows
$f\left(x\right)=x^2+2x+1$ :$-7\le x<-5$
$x+5$ ; $-5\le x\le 2$
$x-1$ ;$2<x<6$
Find :$\left(a\right)f(-7)-f(-3)$

$\left(b\right)\frac{4f(-3)+2f\left(4\right)}{f(-6)-3f\left(1\right)}$

Answer 1

$f\left(x\right)=x^2+2x+1$ for $-7\le x\le -5$

$\Rightarrow f(-7)={(-7)}^2+2\times -7+1=49-14+1=36$

and $f(-6)={(-6)}^2+2\times -6+1=36-12+1=25$

Consider $f\left(x\right)=x+5$ for $-5\le x\le 2$

$f(-3)=-3+5=2$

and $f\left(1\right)=1+5=6$

Consider $x-1$ for $2<x<6$

$f\left(4\right)=4-1=3$

Substituting the above values in

$a)f(-7)-f(-3)=36-2=34$

$b)\frac{4f(-3)+2f\left(4\right)}{f(-6)-3f\left(1\right)}$

$=\frac{4\times 2+2\times 3}{25-3\times 6}=\frac{8+6}{25-18}=\frac{14}{7}=2$