Trying to review for my test next week and this is an example problem that I have a feeling will be on the test yet I am somewhat lost on how to solve. Thanks in advance for the help!
Find the locations of all absolute extrema for the function....
f(x) = 2x^3 + 3x^2 -36x +12 with the domain [ -2, 4 ]
I know I'm supposed to use the first derivative test correct?
f'(x) = 6x^2 + 6x - 36
0 = 6x^2 + 6x - 36
Could someone please solve it from here(assuming I am right so far)?
Three points:Originally Posted by Mathamateur
1) You are looking for "relative" extrema within the interval. This function has no absolute extrema.
2) Don't forget the end points of the domain!
3) Dude!This is a quadratic. If you can't figure out how to solve it, just use the quadratic formula. So saying:
So x = 2 or -3.
-Dan
It says specifiacally to look for any "absolute" extrema though.
So should I plug both of those numbers back into the original to find where they are maximums and minumums. Should I also plug the 2 numbers in the domain in? If so, how do I know when they are "absolute" extrema?
If someone could show me how to do this I would be very thankful.
Poor use of language it means find the global extrema in [-2, 4].
Topsquark has forun the relative extrema of f(x) occur at x=2, -3. Now -3
is not in the interval so is not what we want.
Hence the global maximum in [-2,4] occurs at one of the points x=-2, x=2 or x=4, and the global minimum is also at one of these.
Now f(x) takes values 80, -32, 44 at these points, so the global maximum is 80, and the global minimum is -32.
RonL
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