How do you find the value of c guaranteed by the mean value theorem if it can be applied for #f(x)=x^(1/3)# in the interval [-5,4]?

Question

Evelyn Agard · Accepted Answer

Since #f'(x) = 1/(3 root(3)x^2)# is not defined at #0# (which is in #(-5,4)#), the function is not differentiable on the interval #(-5,4)#, so we cannot use the mean value theorem. The mean value theorem cannot be applied to this function on this interval. Interesting Note:
Although we cannot use the mean value theorem to deduce its existence, there is, in fact, a value of #c# in the interval #(-5,4)# with #f'(c)=(f(4)-f(-5))/(4-(-5))#. #c = sqrt(3/(root(3)4+root(3)5))^3# which is about #0.8678# will work. (Yes, I know I can rationalize the denominator of #3/(root(3)4+root(3)5)#, but that's not what this note is about.)

Isabella Ackerman · Answer

undefined To find the value of $ c $ guaranteed by the Mean Value Theorem for $ f(x) = x^{1/3} $ on the interval $[-5, 4]$, you need to follow these steps:

1. Find the derivative of $ f(x) $ using the Power Rule: $ f'(x) = \frac{1}{3}x^{-\frac{2}{3}} $.

2. Evaluate $ f'(x) $ at the endpoints of the interval: $ f'(-5) = \frac{1}{3}(-5)^{-\frac{2}{3}} $ and $ f'(4) = \frac{1}{3}(4)^{-\frac{2}{3}} $.

3. Calculate the average rate of change of $ f(x) $ over the interval: $ \frac{f(4) - f(-5)}{4 - (-5)} $.

4. Use the Mean Value Theorem formula: $ f'(c) = \frac{f(b) - f(a)}{b - a} $, where $ a = -5 $, $ b = 4 $, and $ c $ is the value we're looking for.

5. Solve for $ c $ by equating $ f'(c) $ to the average rate of change calculated in step 3.

6. Simplify and solve for $ c $ to find the specific value guaranteed by the Mean Value Theorem.

Here are the computations:

- $ f'(x) = \frac{1}{3}x^{-\frac{2}{3}} $
- $ f'(-5) = \frac{1}{3}(-5)^{-\frac{2}{3}} = \frac{1}{3} \times \frac{1}{\sqrt[3]{25}} $
- $ f'(4) = \frac{1}{3}(4)^{-\frac{2}{3}} = \frac{1}{3} \times \frac{1}{\sqrt[3]{16}} $
- Average rate of change $ = \frac{f(4) - f(-5)}{4 - (-5)} $
- $ = \frac{4^{\frac{1}{3}} - (-5)^{\frac{1}{3}}}{9} $
- Mean Value Theorem: $ f'(c) = \frac{f(4) - f(-5)}{4 - (-5)} $
- $ \frac{1}{3}c^{-\frac{2}{3}} = \frac{4^{\frac{1}{3}} - (-5)^{\frac{1}{3}}}{9} $
- Solve for $ c $: $ c^{-\frac{2}{3}} = \frac{3(4^{\frac{1}{3}} - (-5)^{\frac{1}{3}})}{9} $
- $ c = \left(\frac{3(4^{\frac{1}{3}} - (-5)^{\frac{1}{3}})}{9}\right)^{-\frac{3}{2}} $

This gives you the specific value of $ c $ guaranteed by the Mean Value Theorem for $ f(x) = x^{1/3} $ on the interval $[-5, 4]$.