finding absolute maximum and absolute minimum

Nov 2009
18
0
I need to find the absolute maximum and minimum of the function below on the interval [0,4]
g(x) = x / x^2 + 4

I know that i need to find g' which i think is
g'(x) = (x^2 + 4)1 - x(2x)
(x^2 + 4)^2
g'(x) = 4 - x^2
(x^2 - 4)^2

Am I correct so far? Do I need to do g'' or should I factor out and find the critical values?
 
Sep 2008
5,237
1,625
Melbourne
I get

\(\displaystyle g(x) = \frac{x}{x^2+4}\)

\(\displaystyle g'(x) = \frac{(x^2+4)\times 1 - 2x \times x}{(x^2+4)^2}\)

\(\displaystyle g'(x) = \frac{x^2+4 - 2x^2}{(x^2+4)^2}\)

\(\displaystyle g'(x) = \frac{4 - x^2}{(x^2+4)^2}\)

making g'(x)=0

\(\displaystyle 0 = \frac{4 - x^2}{(x^2+4)^2}\)

\(\displaystyle 0 = 4 - x^2\)

\(\displaystyle x=\pm 2\)

You can sub these values into g''(x) if you want to know if they are mins or maxs. Or if you know the general shape of the orignal function it shouldn't be required.