# Stuck on absolute extrema problem

• Mar 22nd 2009, 09:17 PM
Jim Marnell
Stuck on absolute extrema problem
Find locations (ordered pairs) of all absolute extrema for the following function with the specific domain.

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

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

Thanks!
• Mar 22nd 2009, 09:43 PM
Chris L T521
Quote:

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

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

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

Thanks!

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

Since $\displaystyle f^{\prime}\!\left(x\right)$ never equals zero, we must consider where it is undefined. This is when $\displaystyle x+3=0\implies x=-3$. Since $\displaystyle -3\notin\left[0,3\right]$ 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:

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

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

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

Does this make sense?
• Mar 23rd 2009, 05:36 AM
Jim Marnell
Thanks, I understand how to figure these problems out now!