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If f(x) and g(x) are functions derivable in [a,b] such that f(a) = 4, f(b) = 10, g(a) =1, g(b) =3.Show that for a < c < b, we have $ f'(c)=3g'(c).$

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  • If a function is differentiable in an interval it is continuous in that interval.
  • As per L.M.V Theorem if a function is continuous and differentiable then $ \exists $ atleast point $ c \in ( a, b )$ such that $ f'(c) = \large\frac{f(b)-f(a)}{b-a}$
Step 1
From L.M.V theorem
$ f' (c)=\large\frac{10-4}{b-a}\: and \: g' (c)=\large\frac{3-1}{b-a}$
$ f' (c) =\large \frac{6}{b-a} \: and \: \large\frac{2}{b-a} = g' (c)$
Step 2
$ \Rightarrow f' (c) = 3 g' (c)$


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

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