Finding the maximum and minimum

**Truthbetold** · January 9th 2008, 05:47 PM

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.

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

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

These are only 2 critical points; I need 4.

Thanks for the help!

**Truthbetold** · January 9th 2008, 06:16 PM

Derivative is $x^{2/3} + \frac{2(x+2)}{3x^(1/3)}$

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

What did I do wrong?

**topsquark** · January 9th 2008, 07:25 PM

Originally Posted by Truthbetold

Derivative is $x^[2/3] + \frac{2(x+2)}{3x^(1/3)}$

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

What did I do wrong?

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

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

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

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

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