# Thread: Stuck on absolute extrema problem

1. ## Stuck on absolute extrema problem

Find locations (ordered pairs) of all absolute extrema for the following function with the specific domain.

$f(x)= \frac{1-x}{3+x}$ Domain: [0,3]
$f'(x)=\frac{(3+x)(-1)-(1-x)(1)}{(3+x)^2}$

What do i do after this to find the absolute extrema (min,max)

Thanks!

2. Originally Posted by Jim Marnell
Find locations (ordered pairs) of all absolute extrema for the following function with the specific domain.

$f(x)= \frac{1-x}{3+x}$ Domain: [0,3]
$f'(x)=\frac{(3+x)(-1)-(1-x)(1)}{(3+x)^2}$

What do i do after this to find the absolute extrema (min,max)

Thanks!
Note that your derivative simplifies to $f^{\prime}\!\left(x\right)=-\frac{4}{\left(x+3\right)^2}$

Since $f^{\prime}\!\left(x\right)$ never equals zero, we must consider where it is undefined. This is when $x+3=0\implies x=-3$. Since $-3\notin\left[0,3\right]$ we can't consider it to be a critical point.

Since the interval of interest is a closed interval, consider the value of the function at the endpoints:

$f\!\left(0\right)=\frac{1}{3}$

$f\!\left(3\right)=-\frac{2}{9}$

Thus, the absolute maximum is at $\left(0,\tfrac{1}{3}\right)$ and the absolute minimum is at $\left(3,-\tfrac{2}{9}\right)$.

Does this make sense?

3. Thanks, I understand how to figure these problems out now!