f:[2,10] rightarrow R is defined as f(x) = be | vedclass

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(C) First, calculate the values at the endpoints: $f(2) = \frac{1}{2}(2-6)^2 - 3 = \frac{1}{2}(16) - 3 = 8 - 3 = 5$. $f(10) = 10 - 5 = 5$. Since $f(2)

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Rolle's theorem is not applicable for $f(x)$ in $[2,10]$.

Vedclass · Answer

(C) First, calculate the values at the endpoints: $f(2) = \frac{1}{2}(2-6)^2 - 3 = \frac{1}{2}(16) - 3 = 8 - 3 = 5$. $f(10) = 10 - 5 = 5$. Since $f(2) = f(10) = 5$, the first condition of Rolle's theorem is satisfied. Next, check for continuity at $x=4$: Left-hand limit: $\lim_{x 	o 4^-} f(x) = \frac{1}{2}(4-6)^2 - 3 = \frac{1}{2}(4) - 3 = 2 - 3 = -1$. Right-hand limit: $\lim_{x 	o 4^+} f(x) = 4 - 5 = -1$. Since $f(4) = -1$, the function is continuous at $x=4$. Finally, check for differentiability at $x=4$: Left-hand derivative: $f'(x) = (x-6)$, so $f'(4^-) = 4-6 = -2$. Right-hand derivative: $f'(x) = 1$, so $f'(4^+) = 1$. Since $f'(4^-) 
eq f'(4^+)$, the function is not differentiable at $x=4$. Because the function is not differentiable on the interval $(2, 10)$, Rolle's theorem is not applicable.

$f:[2,10] \rightarrow R$ is defined as $f(x) = \begin{cases} \frac{1}{2}(x-6)^2-3, & x \leq 4 \\ x-5, & x > 4 \end{cases}$. Which of the following is true?

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