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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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)

Vedclass · Accepted Answer

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?

Similar Questions

Let $f(x)$ be a differentiable function in $[0, 2]$,$f(0) = 0$ and $f'(x) \le \frac{1}{2}$ for all $x \in [0, 2]$. Then:

If $f(x) = \log(\sin x)$,$x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$,then the value of $c$ by applying Lagrange's Mean Value Theorem $(LMVT)$ is:

If the function $f(x) = x(x + 3) e^{-x/2}$ satisfies Rolle's theorem in the interval $[-3, 0]$,then find the value of $c$.

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

Let $f$ be continuous on $[1, 5]$ and differentiable in $(1, 5).$ If $f(1)=-3$ and $f'(x) \ge 9$ for all $x \in (1, 5)$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module