Suppose that f is differentiable for all x an | vedclass

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

Question

(B) Using $LMVT$ for $f$ in the interval $[1, 2]$:There exists $c \in (1, 2)$ such that $\frac{f(2) - f(1)}{2 - 1} = f'(c)$.Since $f'(c) \le 2$,we hav

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Using $LMVT$ for $f$ in the interval $[1, 2]$:There exists $c \in (1, 2)$ such that $\frac{f(2) - f(1)}{2 - 1} = f'(c)$.Since $f'(c) \le 2$,we have $f(2) - f(1) \le 2(1)$,which implies $f(2) - 2 \le 2$,so $f(2) \le 4$....$(1)$Again,using $LMVT$ for $f$ in the interval $[2, 4]$:There exists $d \in (2, 4)$ such that $\frac{f(4) - f(2)}{4 - 2} = f'(d)$.Since $f'(d) \le 2$,we have $f(4) - f(2) \le 2(2)$,which implies $8 - f(2) \le 4$,so $f(2) \ge 4$....$(2)$From equations $(1)$ and $(2)$,we conclude that $f(2) = 4$.

Suppose that $f$ is differentiable for all $x$ and that $f'(x) \le 2$ for all $x$. If $f(1) = 2$ and $f(4) = 8$,then $f(2)$ has the value equal to

Similar Questions

In the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(c)$. If $a = 4$,$b = 9$ and $f(x) = \sqrt{x}$,then the value of $c$ is:

Let $a > 0$ and $f$ be continuous in $[-a, a]$. Suppose that $f'(x)$ exists and $f'(x) \le 1$ for all $x \in (-a, a)$. If $f(a) = a$ and $f(-a) = -a$,then $f(0)$ is:

If $L.M.V.T.$ is true for $f(x) = x(x-1)(x-2)$ on the interval $x \in [0, 1/2]$,then find the value of $C$.

For the function $f(x)=(x-1)(x-2)$ defined on $\left[0, \frac{1}{2}\right]$,the value of $c$ satisfying Lagrange's mean value theorem is

If $f(x)=|x-2|, x \in[0,4]$ then the Rolle's theorem cannot be applied to the function because

Vedclass Test Series

Exam Paper Generator

Online Exam Module