Finding Absolute Extreme Values of f(x)=2x^3-3x^2-12x+1 [-2,3]

[Sethka]

Hey guys,

f(x)=2x^3-3x^2-12x+1 [-2,3]

I need to use the extreme value theorem to find the absolute extreme values of this function.

So far, I've got:

f'(x)=6x^2-6x-12=6(x^2-x-2)=6(x-2)(x+1)
x= 2,-1

Now what do I do? Please explain in a manner a non-math student can comprehend please.

Thank you!

 

[nocturnal]

Hi Sethka,

The extreme value theorem isn't used to find absolute extrema, rather it tells us that they exist for certain kinds of functions. More specifically it says that continuous functions defined on a closed interval have absolute extrema, but doesn't tell us anything about where they might be. Your function is continuous, as all polynomials are, and is defined on the closed interval [-2,3], so the extreme value theorem guarantees the existence of absolute extrema. This is important, cause you don't want to go and try and find extrema if there aren't any.

Ok, back to the problem. You have found the roots of the derivative of f, ie: values of x for which f'(x) = 0. These along with the endpoints of the interval [-2, 3] are the possible locations for the absolute extrema. What you do now is plug these points into f(x) and find which one makes f(x) the largest and the smallest. Those values of x are the locations of your maxima and minima respectively.

So in short your going to evaluate f(-1), f(2), f(-2), f(3). The largest of these is the absolute maximum, and the smallest is the minimum.

 

[Sethka]

Thanks!

Oh Wow Nocturnal! You've just helped me make sense of something I never really understood for the last two months! You're awesome! Thanks! My textbook sounds belligerent on the topic, so I skipped it and just never really learned it all this time.

 

[nocturnal]

Your welcome. You should also be aware that points in the domain of f, where f'(x) doesn't exist are also candidates for locations of extrema. Together all these points are referred to as "the critical numbers of f."