# Given the function f(x)=((x)/(x+4)) , how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,8]?

Oct 22, 2017

$f \left(x\right) = \frac{x}{x + 4}$ is a rational function, so it is continuous and differentiable where it is defined, which is $\left(- \infty , - 4\right) \cup \left(- 4 , \infty\right)$. Therefore, it satisfies the hypotheses of the Mean Value Theorem on the interval $\left[1 , 8\right] \subseteq \left(- \infty , - 4\right) \cup \left(- 4 , \infty\right)$.

#### Explanation:

The Mean Value Theorem states that if $f : \left[a , b\right] \setminus \rightarrow \mathbb{R}$ is differentiable on $\left(a , b\right)$ and continuous on $\left[a , b\right]$, then there exists a number $c \setminus \in \left(a , b\right)$ such that $f ' \left(c\right) = \frac{f \left(b\right) - f \left(a\right)}{b - a}$ (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 left-most and right-most points on the graph of $f$).

The given function, as mentioned above, 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 $c \setminus \in \left(1 , 8\right)$ such that $f ' \left(c\right) = \frac{f \left(8\right) - f \left(1\right)}{8 - 1} = \frac{\frac{2}{3} - \frac{1}{5}}{7} = \frac{1}{15}$.

It's not necessary to do, but we can also attempt to find the value(s) of $c$ for which this is true. The Quotient Rule implies that

$f ' \left(x\right) = \frac{\left(x + 4\right) \cdot 1 - x \cdot 1}{{\left(x + 4\right)}^{2}} = \frac{4}{{\left(x + 4\right)}^{2}}$.

The value(s) of $c$ therefore satisfy $\frac{4}{{\left(c + 4\right)}^{2}} = \frac{1}{15}$, or ${\left(c + 4\right)}^{2} = 60$. Hence, $c = - 4 \pm \sqrt{60} = - 4 \pm 2 \sqrt{15}$.

Only one of these is in the interval $\left(1 , 8\right)$. So, for this problem, there is one point $c$ where the slope of the tangent line equals the slope of the secant line, $c = - 4 + 2 \sqrt{15} \approx 3.746$.

Oct 22, 2017

See explanation

#### Explanation:

The mean value theorem states that, if f(x) is continuous on an interval [a,b] and differentiable on that same interval, then there is at least one point on that interval (call it c) such that

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

Basically, if you connect points (a,f(a)) and (b,f(b)) with a line, then the slope of that line will be the same as the derivative of f(x) at some point in that interval.

So $f \left(8\right) = \frac{8}{8 + 4} = \frac{8}{12} = \frac{2}{3}$
and $f \left(1\right) = \frac{1}{1 + 4} = \frac{1}{5}$

and $\frac{f \left(b\right) - f \left(a\right)}{b - a} = \frac{\frac{2}{3} - \frac{1}{5}}{7} = \frac{\frac{7}{15}}{7} = \frac{1}{15}$

so next we'll find f'(x). Using the rule for the derivative of the quotient of 2 functions, we have

$f ' \left(x\right) = \frac{\left(x + 4\right) - x}{x + 4} ^ 2$
$f ' \left(x\right) = \frac{4}{x + 4} ^ 2$

Set this equal to $\frac{1}{15}$

$\frac{1}{15} = \frac{4}{{x}^{2} + 8 x + 16}$
...cross multiply:

${x}^{2} + 8 x + 16 = 60$

${x}^{2} + 8 x - 44 = 0$

And now, all we have to do is solve this quadratic equation. It turns out to have roots -11.75 and 3.75.

The first one is NOT in interval [1,8] but the second one is. Therefore, this function satisfies the mean value theorem.

GOOD LUCK