# Math Help - Absolute extrema

1. ## 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.

2. Factoring $2$ from the numerator of $F'(x)$ produces:

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

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

3. Thanks for finding the error.