Question

What is the interval in which the function $f(x)=\sqrt{9-x^2}$ is increasing? (f (x)>0)

• A. 0< x < 3
• B. -3< x < 0
• C. 0 < x < 9
• D. -3 < x < 3

Answer 1

Answer: (B)

$f\left(x\right)=\sqrt{9-x^2}$
$f^{\prime }\left(x\right)=\frac{1}{2\sqrt{9-x^2}}\times \left(-2x\right)=-\frac{x}{\sqrt{9-x^2}}$
For function to be increasing
$-\frac{x}{\sqrt{9-x^2}}>0$
or - x > 0 or x < 0
but $\sqrt{9-x^2}$ is defined only when
$9-x^2>0$ or $x^2-9<0$
(x + 3) (x - 3) < 0
i.e. - 3 < x < 3
- 3 < x < 3 $\cap$ x < 0
$\Rightarrow$ - 3 < x < 0