Question

Let $f\left(x\right)=|9-x^2|$. Then which of the following is true in the interval $-4\le x\le 4$ ?

• A. Maximum value of $f\left(x\right)=7$
• B. Minimum value of $f\left(x\right)=0$
• C. Maximum value of $f\left(x\right)=5$
• D. Minimum value of $f\left(x\right)=5$

Answer 1

Answer: (B)

Minimum value of $f\left(x\right)=0$

$y=f\left(x\right)=x^2-9,x\in [3,\infty )$ or $(-\infty ,-3]$
$y=f\left(x\right)=9-x^2,x\in [-3,3)$
$\therefore$ the graph for f(x) is the branches of the parabola $y+9=x^2$
$x\in (-\infty )-3]\cup [3,\infty )$
and the branch of $x^2=-(y-9),xϵ(-3,3)$
From the figure , the maximum value of $f\left(x\right)is9,atx=0$
$f(-4)=f\left(4\right)=4^2-9=7$
Minimum value $=f(-3)=f\left(3\right)=0$