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Nesterboy [21]
3 years ago
7

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 inte

rval [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.
Mathematics
2 answers:
wlad13 [49]3 years ago
7 0

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}

gogolik [260]3 years ago
4 0

Answer:

x= 16/9

Step-by-step explanation:

f(x) = 2\sqrt{x} is differentiable on [0,4] so the Mean Value Theorem For <u>Integrals</u> 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 <u>average value</u> 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

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