Using Rolle's theorem, find points on the curve

Solution:

The equation of the curve is 

$y=16-x^{2}$    ....(1)

Let $\mathrm{P}\left(x_{1}, y_{1}\right)$ be a point on it where the tangent is parallel to the $x$-axis.

Then,

$\left(\frac{d y}{d x}\right)_{P}=0$         ......(2)

Differentiating (1) with respect to $x$, we get

$\frac{d y}{d x}=-2 x$

$\Rightarrow\left(\frac{d y}{d x}\right)_{P}=-2 x_{1}$

$\Rightarrow-2 x_{1}=0 \quad($ from $(2))$

$\Rightarrow x_{1}=0$

$P\left(x_{1}, y_{1}\right)$ lies on the curve $y=16-x^{2}$

$\therefore y_{1}=16-x_{1}^{2}$

When $x_{1}=0$

$y_{1}=16$

Hence, $(0,16)$ is the required point.