Question

Using LMV Theorem, find a point on the curve $y={(x-3)}^2$, where the tangent is parallel to the chord joining $(3,0)$ and $(5,4)$.

Answer 1

$y=f\left(x\right)={(x-3)}^2$ is continuous and differentiable everywhere.

Slope of tangent to $y=f\left(x\right)$ at any point $=\frac{dy}{dx}=2(x-3)$

$a=3,b=5$

$f\left(a\right)=0,f\left(b\right)=4$

Slope of tangent$=$Slope of chord

$\Rightarrow 2(x-3)=\frac{4}{2}=2$

$\Rightarrow 2x-6=2$

$\Rightarrow 2x=2+6=8$

$\Rightarrow x=\frac{8}{2}=4\in (3,5)$

$\therefore$ the point is $(4,1)$