Answer:
Given,\[f\,(a)=4,\]\[f\,(b)=10,\]\[g\,(a)=1\]and Where, \[a<c<b\] We know that from Lagrange?s mean value theorem, if a function is continuous and differentiable in (a, b), then there exist atleast point \[c\in (a,b)\]such that \[f'(c)=\frac{f\,(b)-f\,(a)}{b-a}\] For the given question \[f'(c)=\frac{10-4}{b-a}=\frac{6}{b-a}\] And \[g'(c)=\frac{3-1}{b-a}=\frac{2}{b-a}\] Here, \[f'(c)=3g'(c)\] Hence proved.
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