Question

How do you determine all values of c that satisfy the mean value theorem on the interval [0, 1] for $ f\left( x \right)=\dfrac{x}{x+6}$ ?

Answer 1

Hint: We should solve these questions by first finding out the derivative of f(x). After getting the derivative, we then have to remove all the values which do not lie in the region of 0 and 1. Then you get your final answer.

Complete step by step answer:
According to the problem, we are asked to determine all values of c that satisfy the mean value theorem on the interval [0, 1] for $ f\left( x \right)=\dfrac{x}{x+6}$
We take the equation as equation 1.
$ f\left( x \right)=\dfrac{x}{x+6}$--- ( 1 )
Now, we find the derivative of f(x) :
$ \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{6}{{{\left( x+6 \right)}^{2}}}$ ------(2)
But according to mean value theorem, we have
$ \begin{align}
  & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{f\left( 1 \right)-f\left( 0 \right)}{1-0} \\
 & \\
\end{align}$------(3)
Now, we find the values of f(!) and f(2).
$ \Rightarrow f\left( 1 \right)=\dfrac{1}{1+6}=\dfrac{1}{7}$ ------ (4)
$ \Rightarrow f\left( 0 \right)=\dfrac{0}{0+6}=0$ ------(5)
Now we substitute the values of equations 4 and 5 in equation number 3. After substituting, we get:
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{\dfrac{1}{7}-0}{1-0}=\dfrac{1}{7}$
Therefore equating 2 and 3, we get,
$ \Rightarrow \dfrac{6}{{{\left( x+6 \right)}^{2}}}=\dfrac{1}{7}$
$ \Rightarrow {\left( x+6 \right)}^{2}=42$
$ \Rightarrow { x+6}=\pm{6.480741}$
$ \Rightarrow { x+6}=+{6.480741} \Rightarrow { x}={0.480741}$
$ \Rightarrow { x+6}=-{6.480741} \Rightarrow { x}={-12.480741 }$
After solving this, we get x = -12.480741 and x = 0.480741. In this both only x = 0.480741 is in the range of [0,1]. There it is the only c that satisfies the mean value theorem.
Therefore, the value c that satisfies the mean value theorem of the given equation $ f\left( x \right)=\dfrac{x}{x+6}$ is x = 0.480741.

Note:
You should be careful in the computations of the values in this question. Also, you should use the $\dfrac{u}{v}$ theorem while calculating the derivative in the above question. You can verify the derivative by integrating it.