question_answer

If f(x) and g(x) are two functions derivable in [a, b], such that f(a) = 4, f(b) = 10, g(a) = 1 and g(b) = 3,  show that for a<c<b, f?(c) = 3g?(c.)

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.