Given the function # f(x) = 9/x^3#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,3] and find the c?

Therefore this function satisfies the hypotheses of the Mean Value Theorem on this interval.

To determine if $f(x) = \frac{9}{x^3}$ satisfies the hypotheses of the Mean Value Theorem on the interval $[1, 3]$, we need to verify if the function is continuous on the interval $[1, 3]$ and differentiable on the interval $(1, 3)$.

1. Continuity: $f(x) = \frac{9}{x^3}$ is continuous on the interval $[1, 3]$ because it is a rational function, and the denominator $x^3$ does not equal zero on this interval.

2. Differentiability: $f(x) = \frac{9}{x^3}$ is differentiable on the interval $(1, 3)$ because it is a composition of continuous functions, and its derivative $f'(x) = -\frac{27}{x^4}$ is defined and continuous on $(1, 3)$ except at $x = 0$.

Since $f(x) = \frac{9}{x^3}$ satisfies both conditions, it satisfies the hypotheses of the Mean Value Theorem on the interval $[1, 3]$.

To find the value $c$ guaranteed by the Mean Value Theorem, we use the formula:

$f'(c) = \frac{f(3) - f(1)}{3 - 1}$

First, find $f(3)$ and $f(1)$:

$f(3) = \frac{9}{3^3} = \frac{9}{27} = \frac{1}{3}$ $f(1) = \frac{9}{1^3} = 9$

Now, find $f'(c)$ by finding the derivative of $f(x)$ and solving for $c$:

$f'(x) = -\frac{27}{x^4}$

$f'(c) = -\frac{27}{c^4}$

Now, set $f'(c)$ equal to the average rate of change:

$-\frac{27}{c^4} = \frac{\frac{1}{3} - 9}{3 - 1} = \frac{\frac{1}{3} - 9}{2} = \frac{-26\frac{2}{3}}{2} = -\frac{26}{3}$

Solve for $c$:

$-\frac{27}{c^4} = -\frac{26}{3}$ $c^4 = \frac{27}{26}$ $c = \sqrt[4]{\frac{27}{26}}$

So, the value of $c$ guaranteed by the Mean Value Theorem lies in the interval $(1, 3)$ and is approximately $\sqrt[4]{\frac{27}{26}}$.