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.

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LammettHash LammettHash · Answer 1 · 2019-05-21T20:31:26+02:00

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

[tex]f'(x)=\dfrac1{\sqrt x}[/tex]

so that by the MVT, there is some [tex]c\in(0,4)[/tex] such that

[tex]f'(c)=\dfrac{f(4)-f(0)}{4-0}\implies\dfrac1{\sqrt c}=\dfrac{2\sqrt4}4=1[/tex]

[tex]\implies1=\sqrt c\implies \boxed{c=1}[/tex]

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powerover9001 powerover9001 · Answer 2 · 2020-09-16T06:36:31+02:00

Answer:

x= 16/9

Step-by-step explanation:

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

Average Value of the Integral:

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

After evaluating you will get [tex]\frac{8}{3}[/tex]

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

[tex]\frac{8}{3} =2\sqrt{x}\\\\\frac{4}{3} = \sqrt{x}\\\\\frac{16}{9} =x[/tex]