Question

At what points on the following curves, is the tangent parallel to x-axis?
(i) y = x² on [−2, 2]
(ii) y = $e^{1-x^2}$ on [−1, 1]
(iii) y = 12 (x + 1) (x − 2) on [−1, 2].

Answer 1

(i) Let $f\left(x\right)=x^2$
Since $f\left(x\right)$ is a polynomial function, it is continuous on $\left[-2,2\right]$ and differentiable on $\left(-2,2\right)$.

Also, $f\left(2\right)=f\left(-2\right)=4$

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c$\in \left(-2,2\right)$ for which $f\text{'}\left(c\right)=0$.

But $f\text{'}\left(c\right)=0\Rightarrow 2c=0\Rightarrow c=0$

$\therefore f\left(c\right)=f\left(0\right)=0$

By the geometrical interpretation of Rolle's theorem, $\left(0,0\right)$ is the point on $y=x^2$, where the tangent is parallel to the x-axis.

(ii) Let $f\left(x\right)=e^{1-x^2}$
Since $f\left(x\right)$ is an exponential function, which is continuous and derivable on its domain, $f\left(x\right)$ is continuous on $\left[-1,1\right]$ and differentiable on $\left(-1,1\right)$.

Also, $f\left(1\right)=f\left(-1\right)=1$

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c$\in \left(-1,1\right)$ for which $f\text{'}\left(c\right)=0$.

But $f\text{'}\left(c\right)=0\Rightarrow -2c{e}^{1-c^2}=0\Rightarrow c=0\left(\because e^{1-c^2}\ne 0\right)$

$\therefore f\left(c\right)=f\left(0\right)=e$

By the geometrical interpretation of Rolle's theorem, $\left(0,e\right)$ is the point on $y=e^{1-x^2}$ where the tangent is parallel to the x-axis.

(iii) Let $f\left(x\right)=12\left(x+1\right)\left(x-2\right)$ ...(1)

$\Rightarrow$$f\left(x\right)=12\left(x^2-x-2\right)$
$\Rightarrow$$f\left(x\right)=12x^2-12x-24$

Since $f\left(x\right)$ is a polynomial function, $f\left(x\right)$ is continuous on $\left[-1,2\right]$ and differentiable on $\left(-1,2\right)$.

Also, $f\left(2\right)=f\left(-1\right)=0$

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c$\in \left(-1,2\right)$ for which $f\text{'}\left(c\right)=0$.

But $f\text{'}\left(c\right)=0\Rightarrow 24c-12=0\Rightarrow c=\frac{1}{2}$

$\therefore f\left(c\right)=f\left(\frac{1}{2}\right)=-12\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)=-27$ (using (1))

By the geometrical interpretation of Rolle's theorem, $\left(\frac{1}{2},-27\right)$ is the point on $y=12\left(x+1\right)\left(x-2\right)$​ where the tangent is parallel to the x-axis.