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• Notes

Let $f(x)=\frac{1}{4} x^{2}+1 .$ Show that $f(x)$ satisfies the conditions of MUT on $[-1,4]$ and find $C$ that satisfies
the conclusion of the theorem.

$f(x)=\frac{1}{4} x^{2}+1 \quad[-1,4]$

(1) $f(x)$ is Cont on $R$

\ $f(x)$ is cont on $R[-1,4]$

(2) $f^{\prime}(x)=\frac{1}{2} x$

\ F is differentiable on $(-1,4)$

by mvt; $\exists c \in(-1,4)$ such that $f^{\prime}(c)=\frac{f(4)-f(-1)}{4-(-1)}$

$\frac{1}{2} c=\frac{1 / 4(4)^{2}+1-\left(\frac{1}{4}(-1)^{2}+1\right)}{5}$

$c=3 / 2$

Check if $f(x)$ satisfies the MVT where $f(x)=\sqrt[3]{x^{2}},[-8,8]$

(1) $f(x)$ is Cont on $R$

\ $f(x)$ is cont on $[-8,8]$

(2) $f(x)=x^{\frac{2}{3}} \rightarrow f^{\prime}(x)=\frac{2}{3} x^{\frac{2}{3}-1}=\frac{2}{3} x^{-\frac{1}{3}}=\frac{2}{3 \sqrt[3]{x}}$

at $x=0 \longrightarrow f^{\prime}(0)=\frac{2}{3 \sqrt[3]{0}}=\frac{2}{0}$ undefined

\ $f(x)$ is not differentlable on $(-8,8)$