Question

If $f\left(x\right)=x+2,g\left(x\right)=x^2-x-2,$ then find $\frac{g\left(1\right)+g\left(2\right)+g\left(3\right)}{f(-4)+f(-2)+f\left(2\right)}$

Answer 1

We have $f\left(x\right)=x+2$

$\Rightarrow f(-4)=-4+2=-2$

$\Rightarrow f(-2)=-2+2=0$

$\Rightarrow f\left(2\right)=2+2=4$

$g\left(x\right)=x^2-x-2$

$g\left(1\right)=1^2-1-2=1-1-2=-2$

$g\left(2\right)=2^2-2-2=4-2-2=0$

$g\left(3\right)=3^2-3-2=9-3-2=4$

Now substituting the above values in

$\frac{g\left(1\right)+g\left(2\right)+g\left(3\right)}{f(-4)+f(-2)+f\left(2\right)}$ we get

$\frac{g\left(1\right)+g\left(2\right)+g\left(3\right)}{f(-4)+f(-2)+f\left(2\right)}=\frac{-2+0-2}{-2+0+4}=\frac{-4}{2}=-2$