# 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).

Toolbox:
• Let $f(x)$ be a function defined on $[a,b]$ such that
• (i) it is continuous on $[a,b]$
• (ii) it is differentiable on $(a,b)$
• Then,there exists $c\in (a,b)$ such that $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
$y=(x-3)^2$------(1)
The given points are $(3,0)$ and $(4,1)$.
Let $y=f(x)$
Then $f(x)=(x-3)^2,x\in[3,4]$
Since $f(x)$ is a polynomial function $f(x)$ is continuous in $[3,4]$
$f'(x)=2(x-3)$ 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.
Step 2:
$f(3)=0$
$f(4)=1$
$f'(c)=2(c-3)$
Therefore $2c-6=\large\frac{1-0}{4-3}$
$\Rightarrow 2c-6=1$
Therefore $c=\large\frac{7}{2}$
$\large\frac{7}{2}$$\in (3,4) Since c is the x-coordinate of that point at which the tangent is parallel to the chord joining the points (3,0) and (4,1). Therefore putting x=\large\frac{7}{2} in equ(1) we get, y=\big(\large\frac{7}{2}-$$3)^2$
$\;\;=\large\frac{1}{4}$
Hence the required points are $\big(\large\frac{7}{2},\frac{1}{4})$