Question

Find the maximum value of $g\left(f\right(x\left)\right)$ if:
$f\left(x\right)=x+4$ and

$g\left(x\right)=6-x^2$

• A. $-6$
• B. $-4$
• C. $2$
• D. $4$
• E. $6$

Answer 1

The correct option is E $6$

Given, $f\left(x\right)=x+4,g\left(x\right)=6-x^2$
$\therefore g\left(f\right(x\left)\right)=g(x+4)=6-{(x+4)}^2$
The minimum value of ${(x+4)}^2$ is $0$ which occurs at $x=-4$.
So, the maximum value of $g\left(f\right(x\left)\right)$ is $6-0=6$.