# 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).

Toolbox:
• M.V.T is
• Check whether the function is continuous in [ a, b ] and differentiable in ( a, b )
• Then find $c \in ( a, b) / f' (c) = \large\frac{f(b)-f(a)}{b-a}$
• Two lines parallel $\Rightarrow$ their slopes are equal.
Step 1
$f(x)=y=(x-3)^2$  is continuous and differentiable everywhere.
Step 2
Slope of tangent to y = f(x) at any point = $\large\frac{dy}{dx}=2(x-3)$
$a=3\: \: b=5$
$f(a)=0 \: f(b)=4$
Step 3
Slope of tangent = Slope of chord
$\Rightarrow 2(x-3) = \large\frac{4}{2} = 2$
$x=4 \: \in \: (3, 5)$
Step 4
therefore the point is ( 4,1)

edited Apr 4, 2013