# Absolute extrema

• Mar 29th 2011, 03:32 PM
rowdy3
Absolute extrema
Find the absolute extrema if they exist as well as all values of x where they occur, for each function and specified domain.
F(x)= (1-x) / (3+x) ; [0,3]
I did the quotient rule. u= 1-x and u'= 1 and v= 3+x and v'= 1
F'(x)= (3+x)(1) - (1-x)(1) / (3+x)^2
F'(x)= 3+x-1+x / (3+x)^2
F'(x)= 2+2x / (3+x)^2
Would I take out a 2 on top? I need to set it to zero but how do I do that?
0= x/(3+x)^2
The bottom is -3 and undefined. So the only numbers I have to do is 0 and 3. I plug it into the original problem and (0, 1/3) and (3, -1/3). 1/3 is absolute max and -1/3 is absolute min.
• Mar 29th 2011, 04:02 PM
NOX Andrew
Factoring \$\displaystyle 2\$ from the numerator of \$\displaystyle F'(x)\$ produces:

\$\displaystyle F'(x) = 2\dfrac{1 + x}{(3 + x)^2}\$

Consequently, \$\displaystyle F'(x) = 0 \implies x = -1\$. However, \$\displaystyle x = -1\$ is not in the domain of \$\displaystyle F(x)\$, so disregard it.

Therefore, the error in your derivative doesn't change the ultimate answer. Just pointing it out in case you are graded on the correctness of your work.
• Mar 29th 2011, 06:12 PM
rowdy3
Thanks for finding the error.