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Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval. (Round your answer to four decimal places. Enter your answers as a comma-separated list.)
f(x) = 8sqrtx
, [4, 9]

Respuesta :

[tex]\bf f(x)=8\sqrt{x}\iff f(x)=8x^{\frac{1}{2}}\qquad \begin{array}{ccllll} [4&,&9]\\ a&&b \end{array}\\\\ -----------------------------\\\\[/tex]

[tex]\bf \cfrac{dy}{dx}=8\cdot \cfrac{1}{2}x^{-\frac{1}{2}}\implies \cfrac{dy}{dx}=\cfrac{4}{\sqrt{x}}\\\\ -----------------------------\\\\ \textit{now, the mean value theorem says}\qquad f'(c)=\cfrac{f(b)-f(a)}{b-a}\\\\ -----------------------------\\\\ thus\qquad \cfrac{4}{\sqrt{c}}=\cfrac{f(9)-f(4)}{9-4}\implies \cfrac{4}{\sqrt{c}}=\cfrac{24-16}{5}\implies \cfrac{4}{\sqrt{c}}=\cfrac{8}{5} \\\\\\ \cfrac{5}{2}=\sqrt{c}\implies \boxed{\cfrac{25}{4}=c}[/tex]
varshamittal029

The value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval will be 25 / 4.

What is Mean Value Theorem?

The Mean Value Theorem is a continuous function on a closed, bounded interval that has at least one point where it is equal to its average value on the interval.

We have,

f(x) = 8√x  and  [4, 9]

Now,

Using The Mean Value Theorem,

i.e.  f'(c) = [ f(b) - f(a)] / (b - a)

So,

f(x) = 8√x = 8 (x) ¹/² = 8 × (1/2) x ⁻¹/² = 4 / √x

f(x) = 4 / √x

So,

f'(c) =  4 / √c

Now,

For a = 4,

f(x) = 8√x  

f(a) = 8 √4 = 8 × 2 = 16

Now,

For b = 9,

f(x) = 8√x  

f(b) = 8 √9 = 8 × 3 = 24

So,

Using The Mean Value Theorem,

i.e.  f'(c) = [ f(b) - f(a)] / (b - a)

And substituting values ;

we get,

f'(c) =  [ 24 - 16 ] / (9 - 4)

4 / √c = 8 / 5

⇒

√c = 5 / 2

⇒

c = 25 /4

Hence we can say that the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval will be 25 / 4.

To learn more about Mean Value Theorem click here,

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