Question

Given the function $f\left( x \right)=\left( \dfrac{x}{x+4} \right)$, how do you determine whether $f$ satisfies the hypotheses of the Mean Value Theorem on the interval $\left[ 1,8 \right]$?

Answer 1

In order to find the solution to the given question that is how to you determine whether $f\left( x \right)=\left( \dfrac{x}{x+4} \right)$ satisfies the hypotheses of the Mean Value Theorem on the interval $\left[ 1,8 \right]$, check for the conditions of hypotheses of Mean value theorem that are whether given function is differentiable on $\left( 1,8 \right)$, whether the function is continuous on $\left[ 1,8 \right]$ and if there exist $c\in \left( 1,8 \right)$ such that ${f}'\left( c \right)=\dfrac{f\left( 8 \right)-f\left( 1 \right)}{8-1}$.

Complete step-by-step solution:
According to the question, given function in the question is as follows:
$f\left( x \right)=\left( \dfrac{x}{x+4} \right)$
The Mean Value Theorem states that if $f:\left[ a,b \right]\to \mathbb{R}$ is differentiable on $\left( a,b \right)$ and continuous on $\left[ a,b \right]$, then there exists a number $c\in \left( a,b \right)~$ such that ${f}'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$ (there exists a number where the slope of the tangent line to the graph of $f$at that point is equal to the slope of the secant line between the leftmost and rightmost points on the graph of $f$).
The given function $f\left( x \right)=\left( \dfrac{x}{x+4} \right)$, is continuous and differentiable everywhere except at $x=-4$. It is therefore continuous on $\left[ 1,8 \right]$ and differentiable on $\left( 1,8 \right)$. The hypotheses of the Mean Value Theorem are satisfied.
The truth of the Mean Value Theorem thus implies that the conclusion of the Mean Value Theorem will be satisfied for this function on this interval. That is, there will exist a number $c\in \left( 1,8 \right)$ such that
$\Rightarrow {f}'\left( c \right)=\dfrac{f\left( 8 \right)-f\left( 1 \right)}{8-1}$.
$\Rightarrow {f}'\left( c \right)=\dfrac{\dfrac{2}{3}-\dfrac{1}{5}}{7}$
$\Rightarrow {f}'\left( c \right)=\dfrac{1}{15}$
So next we'll find ${f}'\left( x \right)$. Using the rule for the derivative of the quotient of two functions, we have:
$\Rightarrow {f}'\left( x \right)=\dfrac{\left( \left( x+4 \right)-x \right)}{{{\left( x+4 \right)}^{2}}}$
$\Rightarrow {f}'\left( x \right)=\dfrac{4}{{{\left( x+4 \right)}^{2}}}$
Set the above equation equal to $\dfrac{1}{15}$, we get:
$\Rightarrow \dfrac{4}{{{\left( x+4 \right)}^{2}}}=\dfrac{1}{15}$
Now solve the terms in the denominator in the bracket we get:
$\Rightarrow \dfrac{4}{{{x}^{2}}+8x-16}=\dfrac{1}{15}$
After this cross multiply the above equation, we get:
$\Rightarrow {{x}^{2}}+8x-16=60$
$\Rightarrow {{x}^{2}}+8x-44=0$
And now, after solving this quadratic equation we get its roots as $-11.75$ and $3.75$.
The first one is not in interval $\left[ 1,8 \right]$ but the second one is.
Therefore, this function $f\left( x \right)=\left( \dfrac{x}{x+4} \right)$ satisfies the mean value theorem.

Note: Students make mistakes while checking one of the conditions in the hypotheses of the mean value theorem that is function is differentiable at $\left[ a,b \right]$ instead of $\left( a,b \right)$ which is completely wrong leads to the wrong answer.