Let `fa n dg` be differentiable on [0,1] such that `f(0)=2,g(0)=0,f(1)=6a n dg(1)=2.` Show that there exists `c in (0,1)` such that `f^(prime)(c)=2g^(prime)(c)dot`

Answer

Proved

Explanation

Since `f` and `g` are differentiable functions for the interval `[0,1]`

`\therefore` they are also continuous for the interval `[0,1]` Consequently by LMV Theorem,


For `f(x)`,
$$ f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}=\frac{6-2}{1}=4 \quad-(1) $$
For $g(x)$
$$
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