Answer:
[tex]c= \dfrac{9}{4}[/tex]
Step-by-step explanation:
For a function f(x) to satisfy the Mean Value Theorem, it must satisfy the following conditions:
Given [tex]f(x)=\sqrt{x}$ in [0,9][/tex]
1. f(x) is defined and continuous on [0,9] since a polynomial function is continuous everywhere.
2. f(x) is differentiable in the interval (0,9) since a polynomial function is differentiable everywhere.
[tex]f(9)=\sqrt{9}=3\\f(0)=\sqrt{0}=0\\f'(x)=\dfrac{1}{2\sqrt{x} }[/tex]
3.By the mean value theorem:
[tex]f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}\\\\\frac{1}{2\sqrt{c}} = \frac{f(9) - f(0)}{9-0} = \frac{3}{9}\\\\\frac{1}{2\sqrt{c}} = \frac{1}{3}\\\\$Cross multiply$\\\\2\sqrt{c}=3\\\\$Square both sides$\\\\(2\sqrt{c})^2=3^2\\\\4c=9\\\\c= \dfrac{9}{4}[/tex]
