Question

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)=5√x,[4,9]

Answer 1

Answer:

25/4

Step-by-step explanation:

The mean value theorem for integrals for the function f(c) over a given interval [a, b] is expressed as g prime(c) = g(b) - g(a)/b-a. The idea is that there is a value c in between the interval [a, b] for the function given.

Given the function g(x) = 5√x within the interval [4,9]

g prime (c) = g(9) - g(4)/9-4

g(9) = 5√9

g(9) = 5*3 = 15

g(4) = 5√4

g(4) = 5*2 = 10

g prime c) = 15-10/9-4

g prime (c) = 5/5

g prime(c) = 1

So we are to find the number for which g prime (x) = g prime(c)

If g(x) = 5√x = $5x^{1/2}$

g prime (x) = $5/2 \ x^{-1/2}$

g prime (x) = 5/2√x

Since g prime (c) = 1 then;

5/2√x = 1

5 = 2√x

√x = 5/2

x = (5/2)²

x = 25/4

The value of c guaranteed by the mid value theorem is 25/4

Answer 2

The possible value of c for $\mathbf{f(x) = 5\sqrt x\ [4,9]}$ is 6.25

The function is given as:

$\mathbf{f(x) = 5\sqrt x\ [4,9]}$

Calculate f(4) and f(9)

$\mathbf{f(4) = 5\sqrt 4 = 10}$

$\mathbf{f(9) = 5\sqrt 9 = 15}$

Substitute c for x in f(x)

$\mathbf{f(c) = 5\sqrt c }$

Calculate f'(c)

$\mathbf{f'(c) = \frac{f(b) - f(a)}{b - a}}$

So, we have:

$\mathbf{f'(c) = \frac{f(9) - f(4)}{9 - 4}}$

$\mathbf{f'(c) = \frac{f(9) - f(4)}{5}}$

This gives

$\mathbf{f'(c) = \frac{15 -10 }{5}}$

$\mathbf{f'(c) = \frac{5 }{5}}$

$\mathbf{f'(c) = 1}$

Also, we have:

$\mathbf{f'(x) = \frac 52x^{-1/2}}$

Substitute c for x

$\mathbf{f'(c) = \frac 52c^{-1/2}}$

Substitute 1 for f'(c)

$\mathbf{\frac 52c^{-1/2} = 1}$

Multiply through by 2/5

$\mathbf{c^{-1/2} = \frac 25}$

This gives

$\mathbf{c^{1/2} = \frac 52}$

Square both sides

$\mathbf{c = \frac{25}4}$

$\mathbf{c = 6.25}$

Hence, the possible value of c is 6.25

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