Question

Determine if the Mean Value Theorem for Integrals applies to the function f of x equals 2 times the square root of x on the interval [0, 4]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.
No, the theorem does not apply.
Yes, x = 1.
Yes, x equals 8 over 3.
Yes, x equals sixteen divided by 9.

Answer 1

$f(x)=2\sqrt x$ is continuous on [0, 4] and differentiable on (0, 4), so the MVT holds. We have

$f'(x)=\dfrac1{\sqrt x}$

so that by the MVT, there is some $c\in(0,4)$ such that

$f'(c)=\dfrac{f(4)-f(0)}{4-0}\implies\dfrac1{\sqrt c}=\dfrac{2\sqrt4}4=1$

$\implies1=\sqrt c\implies \boxed{c=1}$

Answer 2

Answer:

x= 16/9

Step-by-step explanation:

$f(x) = 2\sqrt{x}$ is differentiable on [0,4] so the Mean Value Theorem For Integrals applies.

Average Value of the Integral:

$\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx = \frac{1}{4-0} \int\limits^4_0 {2\sqrt{x} } \, dx$

After evaluating you will get  $\frac{8}{3}$

Now, this is the average value so you still need to find the x-value using the original equation:

$\frac{8}{3} =2\sqrt{x}\\\\\frac{4}{3} = \sqrt{x}\\\\\frac{16}{9} =x$