Given the function #f(x)=x/(x+6)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [0,1] and find the c?

Question

Christopher Allers · Accepted Answer

See below.

You determine whether it satisfies the hypotheses by determining whether #f(x) = x/(x+6)# is continuous on the interval #[0,1]# and differentiable on the interval #(0,1)#. You find the #c# mentioned in the conclusion of the theorem by solving #f'(x) = (f(1)-f(0))/(1-0)# on the interval #(0,1)#. #f# is continuous on its domain, which includes #[0,4]# #f'(x) = 6/(x+6)^2# which exists for all #x != -6# so it exists for all #x# in #(0,1)# Therefore this function satisfies the hypotheses of the Mean Value Theorem on this interval. To find #c# solve the equation #f'(x) = (f(1)-f(0))/(1-0)#. Discard any solutions outside #(0,1)#. You should get #c = -6+sqrt42#.

Aurora Alvarado · Answer

undefined To determine if the function $ f(x) = \frac{x}{x+6} $ satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval $[0,1]$, you need to verify two conditions:

1. The function $ f(x) $ is continuous on the closed interval $[0,1]$.
2. The function $ f(x) $ is differentiable on the open interval $(0,1)$.

First, check for continuity:
- The function $ f(x) $ is continuous for all real numbers except where its denominator equals zero. Since $ x + 6 $ is never zero, $ f(x) $ is continuous everywhere. Hence, it is continuous on the closed interval $[0,1]$.

Next, check for differentiability:
- To check differentiability on the open interval $(0,1)$, differentiate the function $ f(x) $.
  $$ f'(x) = \frac{(x+6) - x}{(x+6)^2} = \frac{6}{(x+6)^2} $$
- Since $ f'(x) $ exists and is continuous for all $ x $ in $(0,1)$, the function $ f(x) $ is differentiable on $(0,1)$.

Since both conditions are satisfied, the Mean Value Theorem can be applied. Now, to find $ c $, use the formula:
$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$
where $ a = 0 $, $ b = 1 $.
$$ f(0) = \frac{0}{0 + 6} = 0 $$
$$ f(1) = \frac{1}{1 + 6} = \frac{1}{7} $$
Now, apply the Mean Value Theorem:
$$ f'(c) = \frac{\frac{1}{7} - 0}{1 - 0} = \frac{1}{7} $$
$$ \frac{6}{(c+6)^2} = \frac{1}{7} $$
$$ 6 = \frac{c+6}{\sqrt{\frac{1}{7}}} $$
$$ c+6 = 6\sqrt{\frac{1}{7}} $$
$$ c = 6\sqrt{\frac{1}{7}} - 6 $$
$$ c \approx -0.66 $$

So, $ c $ approximately equals -0.66.

Adam Adkins · Answer

undefined To determine if the function $f(x) = \frac{x}{x+6}$ satisfies the hypotheses of the Mean Value Theorem on the interval $[0, 1]$, we need to check two conditions:

1. **Continuity:** The function must be continuous on the closed interval $[0, 1]$.
2. **Differentiability:** The function must be differentiable on the open interval $(0, 1)$.

First, let's check for continuity:

1. **Continuity:**  
   The function $f(x)$ is continuous for all real numbers except where the denominator $x + 6$ is equal to zero.  
   However, on the interval $[0, 1]$, the denominator $x + 6$ is never zero. Therefore, $f(x)$ is continuous on $[0, 1]$.

Next, let's check for differentiability:

2. **Differentiability:**  
   For $f(x)$ to be differentiable on $(0, 1)$, it must be differentiable at every point in $(0, 1)$.  
   The derivative of $f(x)$ is given by:  
   $$ f'(x) = \frac{(x+6) - x}{(x+6)^2} = \frac{6}{(x+6)^2} $$  
   This derivative is defined for all $x$ except when $x + 6 = 0$, which is not the case on $(0, 1)$.  
   Therefore, $f(x)$ is differentiable on $(0, 1)$.

Since $f(x)$ is both continuous and differentiable on $[0, 1]$, it satisfies the hypotheses of the Mean Value Theorem on this interval.

To find the $c$ that satisfies the Mean Value Theorem, we use the formula:

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

For the interval $[0, 1]$, $a = 0$ and $b = 1$.  
So, we need to find $c$ such that:

$$ f'(c) = \frac{f(1) - f(0)}{1 - 0} $$

$$ f'(c) = \frac{\frac{1}{1+6} - \frac{0}{0+6}}{1 - 0} $$

$$ f'(c) = \frac{\frac{1}{7} - 0}{1} $$

$$ f'(c) = \frac{1}{7} $$

To find $c$, we need to solve for $c$ in $f'(c) = \frac{1}{7}$.  
Taking the derivative of $f(x)$ again to find $f''(x)$:

$$ f''(x) = -\frac{12}{(x+6)^3} $$

Since $f''(x)$ is negative for all $x$, $f'(x)$ is decreasing on the interval $(0, 1)$.  
Therefore, there is a unique $c$ in $(0, 1)$ such that $f'(c) = \frac{1}{7}$.

The exact value of $c$ can be found by setting $f'(x) = \frac{1}{7}$ and solving for $x$:

$$ \frac{6}{(x+6)^2} = \frac{1}{7} $$

$$ (x+6)^2 = 42 $$

$$ x+6 = \sqrt{42} $$ or $x+6 = -\sqrt{42}$

Since $x$ must be in the interval $(0, 1)$, $x = \sqrt{42} - 6$ is the solution.