How do you determine all values of c that satisfy the conclusion of the mean value theorem on the interval [1,4] for #f(x)=x/(x+9)#?

Question

Ezekiel Alexander · Accepted Answer

Solve #f'(x) = (f(4)-f(1))/(4-1)# on the interval #(1,4)# #9/(x+9)^2 = ((4/13)-(1/10))/(4-1)# #9/(x+9)^2 = 9/130# #x = -9 +-sqrt130# Clearly, #-9-sqrt(130)# is not in the interval, so the only value of #c# that satisfies the conclusion of MVT is #c = -9+sqrt130#.

Emilia Aguilar · Answer

undefined To determine all values of $ c $ that satisfy the conclusion of the Mean Value Theorem on the interval $[1,4]$ for $ f(x) = \frac{x}{x+9} $, you need to follow these steps:

1. Find the average rate of change of $ f(x) $ on the interval $[1,4]$. This is done by calculating $ f(4) - f(1) $ divided by $ 4 - 1 $.
   
2. Next, find the derivative of $ f(x) $, denoted as $ f'(x) $.
   
3. Now, set up the equation according to the conclusion of the Mean Value Theorem: $ f'(c) = \frac{f(4) - f(1)}{4 - 1} $.
   
4. Plug in the values of $ f(4) $ and $ f(1) $ into the equation.
   
5. Solve for $ c $ to find all values that satisfy the conclusion of the Mean Value Theorem.

Let's go through the steps:

1. $ f(4) = \frac{4}{4+9} = \frac{4}{13} $ and $ f(1) = \frac{1}{1+9} = \frac{1}{10} $. So, the average rate of change of $ f(x) $ on $[1,4]$ is:
   
   $$ \frac{f(4) - f(1)}{4 - 1} = \frac{\frac{4}{13} - \frac{1}{10}}{4 - 1} = \frac{4}{13} - \frac{1}{10} $$

2. Now, find the derivative of $ f(x) $:
   
   $$ f'(x) = \frac{d}{dx}\left(\frac{x}{x+9}\right) = \frac{(x+9)(1) - x(1)}{(x+9)^2} = \frac{9}{(x+9)^2} $$
   
3. According to the Mean Value Theorem, $ f'(c) = \frac{4}{13} - \frac{1}{10} $.
   
4. Substitute $ f'(x) $ into the equation:
   
   $$ \frac{9}{(c+9)^2} = \frac{4}{13} - \frac{1}{10} $$
   
5. Solve for $ c $.

$$ \frac{9}{(c+9)^2} = \frac{40}{130} - \frac{13}{130} = \frac{27}{130} $$
   
   $$ (c+9)^2 = \frac{130 \times 9}{27} = 130 \times \frac{9}{27} = 130 \times \frac{1}{3} = \frac{130}{3} $$

$$ c + 9 = \pm \sqrt{\frac{130}{3}} $$
   
   $$ c = -9 \pm \sqrt{\frac{130}{3}} $$

So, the values of $ c $ that satisfy the conclusion of the Mean Value Theorem on the interval $[1,4]$ for $ f(x) = \frac{x}{x+9} $ are $ c = -9 + \sqrt{\frac{130}{3}} $ and $ c = -9 - \sqrt{\frac{130}{3}} $.