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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).

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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)

 

answered Mar 8, 2013 by thanvigandhi_1
edited Apr 4, 2013 by thanvigandhi_1
 

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