Question

The function $f\left(x\right)={(9-x^2)}^2$ increases in

• A. $(-3,0)\cup (3,\infty )$
• B. $(-\infty ,-3)\cup (3,\infty )$
• C. $(-\infty ,-3)\cup (0,3)$
• D. $(-3,3)$
• E. $(3,\infty )$

Answer 1

The correct option is A

$(-3,0)\cup (3,\infty )$

Explanation for correct option

Given function $f\left(x\right)={(9-x^2)}^2$

$\Rightarrow f\text{'}\left(x\right)=2(9–x^2)(-2x)$

The function $f\left(x\right)$ will be increasing if $f\text{'}\left(x\right)>0$

$$\begin{gathered} \Rightarrow 2(9–x^2)(-2x)>0 \\ \Rightarrow 2(3^2–x^2)(-2x)>0 \\ \Rightarrow 2(3-x)(3+x)(-2x)>0 \\ \Rightarrow (3-x)(3+x)(-2x)>0 \end{gathered}$$

Thus, $f\left(x\right)$ is increasing in $x\in (-3,0)\cup (3,\infty )$

Hence, $\mathrm{Option}\left(\mathrm{A}\right)$ is correct.