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:

  • A
    $8$
  • B
    $5.25$
  • C
    $4$
  • D
    $6.25$

Explore More

Similar Questions

If from the Mean Value Theorem,$f'({x_1}) = \frac{f(b) - f(a)}{b - a}$,then

Let $f(x)$ be a differentiable function for all $x$ such that $f'(x) \le 5$ and $f(1) = 4$. The maximum value of $f(5)$ is...

Let $f:(a, b) \rightarrow R$ be a twice differentiable function such that $f(x) = \int_{a}^{x} g(t) \, dt$ for a differentiable function $g(x)$. If $f(x) = 0$ has exactly five distinct roots in $(a, b)$,then $g(x) g'(x) = 0$ has at least:

The equation $2^x+5^x=3^x+4^x$ has

Consider the function $f(x) = |x - 2| + |x - 5|$,$x \in R$.
Statement-$1$: $f'(4) = 0$.
Statement-$2$: $f$ is continuous in $[2, 5]$,differentiable in $(2, 5)$,and $f(2) = f(5)$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo