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Q.
What is the interval in which the function $f(x) = \sqrt{9 - x^2}$ is increasing? (f (x)>0)
Application of Derivatives
Solution:
$f\left(x\right) = \sqrt{9-x^{2}}$
$ f'\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