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

  • $\begingroup$ 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.] $\endgroup$ – colormegone Apr 11 '14 at 2:23
  • $\begingroup$ 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. $\endgroup$ – colormegone Apr 11 '14 at 2:57

Notice that this is the top half of a circle. At the endpoints, the function is vertical, and in the middle the function is flat.'


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