Verify lagranges mean value theorem for the following functions on the indicated intervals. Also, find a point `c` in the indicated interval: `f(x)=x(x-2)on[1,3]` `f(x)=x(x-1)(x-2)on[0,1/2]dot`

Explanation

First Question


`f(x)=x(x-2)` on `[1,3]`


`f(x)` is polynomial function so it is continuous in `[1,3]`


and differentiable in `[1,3]`


So, there must exist at least one real number `c \in(1,3)` such that `f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(1)}{3-1}`


`f^{\prime}(x)=2 x-2 , f^{\prime}(c)=2 c-2, f(3)=9-6=3` and `f(1)=1-2=-1`


`f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \Rightarrow 2 c-2=\frac{3-(-1)}{3-1} \Rightarrow 2 c-2=2 \Rightarrow c=2`


As, `c=2 \in(1,3)` such that `f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}`


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