Question

Find a point on the curve $y={(x-3)}^2$,where the tangent is parallel to the chord joining the point $(3,0)$ and $(4,1)$

Answer 1

$y={(x-3)}^2$ ........$\left(1\right)$

The given points are $(3,0)$ and $(4,1)$

Let $y=f\left(x\right)$

Then $f\left(x\right)={(x-3)}^2$,$x\in [3,4]$

Since $f\left(x\right)$ is a polynomial function $f\left(x\right)$ is continuous in $[3,4]$

$f^{\prime }\left(x\right)=2(x-3)$ which exists in $(3,4)$

Therefore it is also differentiable in $(3,4)$

Hence both the conditions of Lagrange's Mean value theorem is satisfied.

$f\left(3\right)=0$

$f\left(4\right)=1$

$f^{\prime }\left(c\right)=2(c-3)$

$\therefore 2c-6=\frac{1-0}{4-3}$

$\Rightarrow 2c-6=1$

$\Rightarrow 2c=7$

$\therefore c=\frac{7}{2}$

$\frac{7}{2}\in (3,4)$

Since $c$ is the $x-$coordinate of that point at which the tangent is parallel to the chord joining the points $(3,0)$ and $(4,1)$.

$\therefore$ putting $x=\frac{7}{2}$ in eqn$\left(1\right)$ we get

$y={(\frac{7}{2}-3)}^2=\frac{1}{4}$

Hence the required points are $(\frac{7}{2},\frac{1}{4})$