# Finding the maximum and minimum

• Jan 9th 2008, 05:47 PM
Truthbetold
Finding the maximum and minimum
Use analytic methods to find the extreme values of the function on the interval and where they occur.

Extreme values includes local and absolute max and min.

Analytic methods meaning I can't graph anything.

$\displaystyle g(x)= \sin (x + \frac{\pi}{4})$
To find the critical points i need to find when y'= 0.
$\displaystyle g'(x)=\cos(x + \frac{\pi}{4})$

I got the points $\displaystyle \frac{\pi}{4}, \frac{5\pi}{4}$
When plugged into g(x) they yield 1 and -1, respectively.

These are only 2 critical points; I need 4.

Thanks for the help!
• Jan 9th 2008, 06:16 PM
Truthbetold
Another Problem
Derivative is $\displaystyle x^{2/3} + \frac{2(x+2)}{3x^(1/3)}$

The original function is $\displaystyle x^{2/3}(x+2)$
The answer says the critical points are 0 and -4/5.
The points appear to be 0, and -2.

What did I do wrong?
• Jan 9th 2008, 07:25 PM
topsquark
Quote:

Originally Posted by Truthbetold
Derivative is $\displaystyle x^[2/3] + \frac{2(x+2)}{3x^(1/3)}$

The original function is $\displaystyle x^[2/3](x+2)$
The answer says the critical points are 0 and -4/5.
The points appear to be 0, and -2.

What did I do wrong?

$\displaystyle x^[2/3] + \frac{2(x+2)}{3x^(1/3)} = 0$

Multiply both sides by $\displaystyle x^{1/3}$. (Note that x = 0 is not in the domain of the derivative, so we lose nothing by doing this.)

$\displaystyle x + \frac{2}{3}(x + 2) = 0$

So I'm getting that x = -4/5 is a critical point.

x = 0 is defined to be a critical point since the derivative doesn't exist there.

-Dan