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

Answer

N/a

Explanation

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' = (x-4).1+x.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`
` rArr y(0) = y(4)`
Since, conditions of Rolle's theorem are satisfied.
Hence, there exists a point such that
`f(c) = 0` in `(0,4), [:' f(x) = y']`
`rArr 2c - 4 = 0`
`rArr c = 2`
`rArr 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.