# Finding Increasing/Decreasing and absolute/local min/max

I am kinda confused and feel like I am getting more confused as I dive into these math problems. First, I know that to find critical points(which are needed for increase/decrease and min/max) we take the derivative, and wherever it =0 or =undefined that those are a critical point.

So for example, I have a problem $\ f(x)=√(25-x^2) \$ , so $\ f'(x)= \frac{-x}{\sqrt{25-x^2}} \$ .

Ok, so that means we have critical points at 0, -5 and 5 correct?

Now by taking these points on intervals (-5,0), (0,5) and seeing where they increase and decrease we can find what we are looking for correct? So on (-5,0) I would say it is decreasing and on (0,5) it is also decreasing correct? (Now just a question with the above, when we have 3 critical points, is there a certain number of domain intervals to check for inc/dec? Because I have 3 crit points but 2 domain intervals?)

After this I am stuck. I know that if a function is increasing then decreasing, that will be a local max, and if its decreasing then increasing its a local min. and if there is no change in sign, there is no extreme, so since there is no extreme, is there no global min/max?

I am kinda stumped, I really need some help with the concept so I can do more of these problems easily.

Thank you

• Your sign is reversed for the derivative on one of your two intervals: $\ f(x) \$ increases on $\ (-5, 0) \$ [since $\ x \$ is negative there] and decreases on $\ (0,5) \ .$ [Hint: the curve described by this function is a semi-circle above the $\ x-$ axis.] – colormegone Apr 11 '14 at 2:23
• Looking back again, as to your other question, the derivative is undefined at the critical points $\ x = \pm 5 \$ (the curve makes that clear why), and the derivative is zero at $\ x = 0 \ .$ The function is only defined on $\ [-5 , 5 ] \$ , so you do only have the two intervals you indicated to consider. – colormegone Apr 11 '14 at 2:57