How do you find the values of C guaranteed by the Mean Value Theorem for #f(x)= 9/x^3# over [1, 3]?

Question

How do you find the values of C guaranteed by the Mean Value Theorem for #f(x)= 9/x^3# over  [1, 3]?

Charles Arocha · Accepted Answer

In the conclusion of the theorem, #c# is a solution in #(1,3)# to the equation #f'(x) = (f(3)-f(3))/(3-1)#

So write the equation and solve it.

#-27/x^4 = (1/3 - 9)/(3-1) = (-26/3)/2 = -13/3#

So

#13x^4 = 81# and

#x = +- 3/root(4)13#

Obviously, the negative solution is not in the interval, so there is only one value of #c#, namely

#c=3/root(4)13#

You should convince yourself that

#1 < 3/root(4)13 < 3#

Ezekiel Alexander · Answer

undefined To find the values of $C$ guaranteed by the Mean Value Theorem for $f(x) = \frac{9}{x^3}$ over the interval $[1, 3]$, follow these steps:

1. Determine the derivative of $f(x)$.
2. Find the average rate of change of $f(x)$ over the interval $[1, 3]$.
3. Apply the Mean Value Theorem to find the value of $C$.

Let's proceed with the steps:

1. The derivative of $f(x) = \frac{9}{x^3}$ is $f'(x) = -\frac{27}{x^4}$.

2. The average rate of change of $f(x)$ over $[1, 3]$ is:
   $$ \text{Average rate of change} = \frac{f(3) - f(1)}{3 - 1} $$

$$ = \frac{\frac{9}{3^3} - \frac{9}{1^3}}{3 - 1} $$
   $$ = \frac{\frac{9}{27} - 9}{2} $$
   $$ = \frac{\frac{1}{3} - 9}{2} $$
   $$ = \frac{-\frac{26}{3}}{2} $$
   $$ = -\frac{13}{3} $$

3. Apply the Mean Value Theorem:
   The Mean Value Theorem states that there exists at least one $C$ in the interval $[1, 3]$ such that $f'(C) = \text{Average rate of change}$.
   
   Therefore, we need to solve the equation $f'(C) = -\frac{13}{3}$ for $C$.

$$ -\frac{27}{C^4} = -\frac{13}{3} $$

Solve for $C$:
   $$ C^4 = \frac{27 \cdot 3}{13} $$
   $$ C^4 = \frac{81}{13} $$
   $$ C = \sqrt[4]{\frac{81}{13}} $$

$C$ is approximately $1.947$.

So, the value of $C$ guaranteed by the Mean Value Theorem for $f(x) = \frac{9}{x^3}$ over the interval $[1, 3]$ is approximately $1.947$.