Question

Using Mean Value Theorem, find a point on the curve y=(x-3)^2, where the tangent is parallel to the chord joining the points (3,0) and (4,1).

Answer 1

$\frac{dy}{dx} = 2(x - 3)$
but
slope of line joining the points
$= (1 - 0) \div (4 - 3)$
$= 1$
By mean value theorem

$2(x - 3) = 1$
$x = \frac{7}{2}$
$y = \frac{1}{4}$
The required point is
$( \frac{7}{2} \: \frac{1}{4} )$