Question
Using Rolle's theorem, find points on the curve $\text{y}=16-\text{x}^2,\text{x}\in[-1,1],$ where tagent is parallel to x-axis.
✓
Answer
The equation of the curve is,
$\text{y}=16-\text{x}^2\ ....(1)$
Let $P(x_1,y_1)$ be a point on it where the tangent is parallel to x-axis.
Then,
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=0\ ....(2)$
Differentiating (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=-2\text{x}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=-2\text{x}_1$
$\Rightarrow-2\text{x}_1=0$ (from(2))
$\Rightarrow\text{x}_1=0$
$P(x_1, y_1)$ lies on the curve $y = 16 - x^2$
$\therefore\text{y}_1=16-\text{x}_1^2$
When $x_1 = 0,$
$y_1 = 16$
Hence, (0, 16) is the required point.
$\text{y}=16-\text{x}^2\ ....(1)$
Let $P(x_1,y_1)$ be a point on it where the tangent is parallel to x-axis.
Then,
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=0\ ....(2)$
Differentiating (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=-2\text{x}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=-2\text{x}_1$
$\Rightarrow-2\text{x}_1=0$ (from(2))
$\Rightarrow\text{x}_1=0$
$P(x_1, y_1)$ lies on the curve $y = 16 - x^2$
$\therefore\text{y}_1=16-\text{x}_1^2$
When $x_1 = 0,$
$y_1 = 16$
Hence, (0, 16) is the required point.
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