Verify Mean Value Theorem for the function $f(x)=x^{2}$ in the interval $[2,4]$

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The function $f(x)=x^{2}$ is continuous in $[2,4]$ and differentiable in $(2,4)$ as its derivative $f^{\prime}(x)=2 x$ is defined in $(2,4).$

Now, $\quad f(2)=4$ and $f(4)=16 .$ Hence

$\frac{f(b)-f(a)}{b-a}=\frac{16-4}{4-2}=6$

$\mathrm{MVT}$ states that there is a point $c \in(2,4)$ such that $f^{\prime}(c)=6 .$ But $f^{\prime}(x)=2 x$ which implies $c=3 .$ Thus at $c=3 \in(2,4),$ we have $f^{\prime}(c)=6$

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