At what points on the following curves, is the tangent parallel to x-axis?

Solution:

(i) Let $f(x)=x^{2}$

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

Also, $f(2)=f(-2)=4$

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

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

But $f^{\prime}(c)=0 \Rightarrow 2 c=0 \Rightarrow c=0$

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

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

(ii) Let $f(x)=e^{1-x^{2}}$

Since $f(x)$ is an exponential function, which is continuous and derivable on its domain, $f(x)$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$.

Also, $f(1)=f(-1)=1$

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

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

But $f^{\prime}(c)=0 \Rightarrow-2 c e^{1-c^{2}}=0 \Rightarrow c=0 \quad\left(\because e^{1-c^{2}} \neq 0\right)$

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

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

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

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

 

$\Rightarrow f(x)=12 x^{2}-12 x-24$

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

Also, $f(2)=f(-1)=0$

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

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

But $f^{\prime}(c)=0 \Rightarrow 24 c-12=0 \Rightarrow c=\frac{1}{2}$

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

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