Question

Using Rolle's theorem, find points on the curve y = 16 − x², x ∈ [−1, 1], where tangent is parallel to x-axis.

Answer 1

The equation of the curve is
$y=16-x^2$. ...(1)

Let P$\left(x_1,y_1\right)$ be a point on it where the tangent is parallel to the x-axis.

Then,
${\left(\frac{dy}{dx}\right)}_P=0$ ...(2)

Differentiating (1) with respect to x, we get

$$\begin{gathered} \frac{dy}{dx}=-2x \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_P=-2x_1 \\ \Rightarrow -2x_1=0\left(\mathrm{from}\left(2\right)\right) \\ \Rightarrow x_1=0 \end{gathered}$$

$P\left(x_1,y_1\right)$ lies on the curve $y=16-x^2$.

$\therefore$ $y_1=16-{x_1}^2$

When $x_1=0$,
$y_1=16$

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