CALCU-Find the number c guaranteed by the Mean Val

Find the number

c

guaranteed by the Mean Value Theorm for

f(x)=2 \sqrt[3]{x}

on

[1,8]

. That is, find the number

c

, with 1<c<8, such that

f^{\prime}(c)=\frac{f(8)-f(1)}{8-1}

c=

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Solution by Steps

step 1

First, we need to find the derivative of the function

f(x) = 2\sqrt[3]{x}

step 2

The derivative

f'(x)

is found using the power rule:

f'(x) = \frac{d}{dx}(2x^{\frac{1}{3}}) = \frac{2}{3}x^{-\frac{2}{3}}

step 3

Next, we calculate the difference quotient for

f(x)

over the interval [1, 8]:

\frac{f(8)-f(1)}{8-1}

step 4

Substituting the values of

f(x)

gives us

\frac{2\sqrt[3]{8}-2\sqrt[3]{1}}{8-1} = \frac{2(2)-2(1)}{7} = \frac{2}{7}

step 5

Now, we set the derivative equal to the difference quotient:

\frac{2}{3}c^{-\frac{2}{3}} = \frac{2}{7}

step 6

Solving for

c

, we multiply both sides by

\frac{3}{2}

to get

c^{-\frac{2}{3}} = \frac{3}{7}

step 7

Taking both sides to the power of

-\frac{3}{2}

gives us

c = \left(\frac{7}{3}\right)^{-\frac{3}{2}}

step 8

Simplifying the right side, we find

c = \left(\frac{3}{7}\right)^{\frac{3}{2}}

step 9

Finally, we calculate

c

to get

c = \left(\frac{3}{7}\right)^{\frac{3}{2}} \approx 0.962

. However, since we need 1 < c < 8 , we must have made a mistake in our calculation. Let's re-evaluate step 7

step 10

Re-evaluating step 7, we raise both sides to the power of

-\frac{3}{2}

correctly:

c = \left(\frac{7}{3}\right)^{\frac{3}{2}}

step 11

Calculating

c

correctly, we find

c = \left(\frac{7}{3}\right)^{\frac{3}{2}} \approx 4.1833

, which satisfies 1 < c < 8

Answer

c \approx 4.1833

Key Concept

Mean Value Theorem

Explanation

The Mean Value Theorem states that for a continuous function on a closed interval [a, b], there exists some c in (a, b) such that

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

. In this problem, we applied the theorem to find the specific value of c for the given function.

Let

f(x)=3 x^{3}-2 x^{2}+2 x-2

on

[-1,2]

. Find all numbers

c

satisfying the Mean Value Theorem.

Solution by Steps

step 1

First, we need to verify that the function

f(x) = 3x^3 - 2x^2 + 2x - 2

is continuous on the closed interval

[-1, 2]

and differentiable on the open interval

(-1, 2)

step 2

Since

f(x)

is a polynomial, it is continuous and differentiable everywhere, hence it satisfies the conditions of the Mean Value Theorem (MVT) on

[-1, 2]

step 3

According to the MVT, there exists at least one number

c

in the open interval

(-1, 2)

such that

f'(c) = \frac{f(2) - f(-1)}{2 - (-1)}

step 4

Calculate

f(2)

and

f(-1)

:

f(2) = 3(2)^3 - 2(2)^2 + 2(2) - 2

and

f(-1) = 3(-1)^3 - 2(-1)^2 + 2(-1) - 2

step 5

Simplify to find

f(2) = 12

and

f(-1) = -3

step 6

Calculate the difference quotient:

\frac{f(2) - f(-1)}{2 - (-1)} = \frac{12 - (-3)}{2 - (-1)} = \frac{15}{3} = 5

step 7

Find

f'(x)

, the derivative of

f(x)

:

f'(x) = \frac{d}{dx}(3x^3 - 2x^2 + 2x - 2) = 9x^2 - 4x + 2

step 8

Set

f'(c) = 5

:

9c^2 - 4c + 2 = 5

step 9

Solve for

c

:

9c^2 - 4c - 3 = 0

step 10

Factor or use the quadratic formula to find the values of

c

that satisfy the equation

step 11

The quadratic formula gives

c = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(9)(-3)}}{2(9)}

step 12

Simplify to find the values of

c

:

c = \frac{4 \pm \sqrt{16 + 108}}{18} = \frac{4 \pm \sqrt{124}}{18}

step 13

The two possible values for

c

are

c = \frac{4 + \sqrt{124}}{18}

and

c = \frac{4 - \sqrt{124}}{18}

, but only the value in the interval

(-1, 2)

is valid for the MVT

step 14

Check which value of

c

is in the interval

(-1, 2)

step 15

The value

c = \frac{4 - \sqrt{124}}{18}

is outside the interval

(-1, 2)

, so the only valid value is

c = \frac{4 + \sqrt{124}}{18}

Answer

c = \frac{4 + \sqrt{124}}{18}

Key Concept

Mean Value Theorem (MVT)

Explanation

The Mean Value Theorem states that if a function is continuous on a closed interval and differentiable on the open interval, then there is at least one point where the derivative equals the average rate of change over the interval.

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