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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 $$(-8,8)$$

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