Let f:[1,3] rightarrow R be a continuous func | vedclass

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

Question

(B) Given that $f:[1,3] \rightarrow R$ is continuous on $[1,3]$ and differentiable in $(1,3)$ with $f^{\prime}(x)=|f(x)|^{2}+4$. By applying the Lagra

Vedclass · Accepted Answer

$f(3)-f(1)=5$ is false

Vedclass · Answer

(B) Given that $f:[1,3] \rightarrow R$ is continuous on $[1,3]$ and differentiable in $(1,3)$ with $f^{\prime}(x)=|f(x)|^{2}+4$. By applying the Lagrange Mean Value Theorem $(LMVT)$,there exists at least one point $c \in (1,3)$ such that: $\frac{f(3)-f(1)}{3-1} = f^{\prime}(c)$ $\frac{f(3)-f(1)}{2} = |f(c)|^{2} + 4$ Since $|f(c)|^{2} \geq 0$,we have $|f(c)|^{2} + 4 \geq 4$. Therefore,$\frac{f(3)-f(1)}{2} \geq 4$,which implies $f(3)-f(1) \geq 8$. Thus,the statement $f(3)-f(1)=5$ is false.

Let $f:[1,3] \rightarrow R$ be a continuous function that is differentiable in $(1,3)$ and $f^{\prime}(x)=|f(x)|^{2}+4$ for all $x \in(1,3).$ Then,

Similar Questions

The value of $c$ for the Lagrange's mean value theorem for $f(x)=\sqrt{x^2-x}, x \in[1,4]$ is

The value $C$ of the Lagrange's mean value theorem for the function $f(x)=x(x-1)(x-2)$ in the interval $[0, 1/2]$ is

Given $f(x) = 4 - (\frac{1}{2} - x)^{2/3}$,$g(x) = \begin{cases} \frac{\tan([x])}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$,$h(x) = \{x\}$,and $k(x) = 5^{\log_2(x + 3)}$. Then,in the interval $[0, 1]$,Lagrange's Mean Value Theorem is $NOT$ applicable to:

If $f(x)$ satisfies the conditions of Rolle's theorem in $[1, 2]$ and $f(x)$ is continuous in $[1, 2]$,then $\int_1^2 f'(x) dx$ is equal to

For which interval does the function $f(x) = \frac{x^2 - 3x}{x - 1}$ satisfy all the conditions of Rolle's theorem?

Vedclass Test Series

Exam Paper Generator

Online Exam Module