Are There More Solutions for f'(x) in This Extrema Problem?

[arturo_026]

find this problem really though. I did solve for it's derivative and then by EVT tried to find its crititcal points. I get (-π/2) and (π/2) for f'(x)=0.
The problm is that I'm not sure if there are more solutions for f'(x).

Please see function in problem 1 of attachment:


Any help will be greatly apreciated.
Thank you.

 

[berkeman]

find this problem really though. I did solve for it's derivative and then by EVT tried to find its crititcal points. I get (-π/2) and (π/2) for f'(x)=0.
The problm is that I'm not sure if there are more solutions for f'(x).

Please see function in problem 1 of attachment:


Any help will be greatly apreciated.
Thank you.

Welcome to the PF. The problem appears to ask for the extrema of the function, in addition to critical points. What does the function do for x approaching +/- infinity?

Have you tried graphing the function on a graphing calculator to get an intuitive feel for what it looks like?

 

[arturo_026]

Thank you.

I have graphed it, however not in great detail, and for what I see it looks somewhat like the function for y = x³.
f(x) decreases without bound as x aproaches -∞ and increases as x --> ∞.
However, according to EVT the function f attains its absolute extrema at either its endpoints a and b, or in the closed interval (a,b). However I was not given any interval.

So does it mean that the extrema for f(x) does not exist?

 

[slider142]

The EVT deals with functions defined on/restricted to a closed interval, so unless you examine your function on such an interval, EVT doesn't have much to say. Ie., if you were to examine f on a closed interval containing its critical points, EVT says to look for extrema at the endpoints and each critical point and that those are the only possible points for extrema, if existent.
Since you are examining this function on its entire domain, you can go ahead and look at its behavior at its critical points only, since there are no endpoints. Note that it may be beneficial in some problems to also examine asymptotic behavior, which is values the function may approach as the argument increases or decreases without bound. These do not usually count as extrema, however, as the function usually never actually attains those values (One exception is a function which becomes constant above a certain value). Your function at a glance does not exhibit this behavior.
Remember that critical points also pertain to points where the derivative is undefined. This is also not the case for your function, so the only possible extrema are where the derivative is 0. If you have not learned how to use higher derivatives to pinpoint extrema, try the old fashioned method using EVT. Make a small closed interval around a critical point and test one value on either side of the point. Since there is only at most one extremum in the closed interval (at the critical point), you only need these two values to know the full behavior on that interval: it is either a maximum (both points are lower), a minimum (both points are higher), or it is not an extremum (the points increase or decrease).

 

[slider142]

I get (-π/2) and (π/2) for f'(x)=0.

By the way, the values you give as critical points don't seem to agree with the derivative function I got. Have you looked at a graph of the derivative function?

 

[arturo_026]

By the way, the values you give as critical points don't seem to agree with the derivative function I got. Have you looked at a graph of the derivative function?

Well, giving it a better thought, i think a more apropriate solution would be (+- pi/2) + n; n being a natural number. If I'm correct.

 

[slider142]

Well, giving it a better thought, i think a more apropriate solution would be (+- pi/2) + n; n being a natural number. If I'm correct.

Actually, your previous value was correct. I had a sign wrong in my derivative. :)