Using Rolle's theorem, find the point on the curve $y = x(x - 4),x \in [0,4],$ where the tangent is parallel to $X$-axis

We have, $y = x(x - 4),x \in [0,4]$

(i) $y$ is a continuous function since $x(x - 4)$ is a polynomial function.

Hence, $y = x(x - 4)$ is continuous in [0,4] .

(ii) ${y^\prime } = (x - 4) \cdot 1 + x \cdot 1 = 2x - 4$ which exists in (0,4) .

Hence, $y$ is differentiable in (0,4) .

(iii) $y(0) = 0(0 - 4) = 0$

and $y(4) = 4(4 - 4) = 0$

$\Rightarrow$ $y(0) = y(4)$

Sicne, conditions of Rolle's theorem are satisfied.

Hence, there exists a point $c$ such that
${f^\prime }(c) = 0$ in (0,4)

$\Rightarrow$ $2c - 4 = 0$

$\Rightarrow$ $c = 2$

$\Rightarrow$ $x = 2;y = 2(2 - 4) = - 4$

Thus, (2,-4) is the point on the curve at which the tangent drawn is parallel to $x$-axis.

Verify mean value theorem for each of the functions.