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Maxima-minima problems

by Kamataat
Tags: maximaminima
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Kamataat
#1
Dec17-04, 11:15 AM
P: 137
Hi!

For example y=-x^3-3x=0 gives y'=-3x^2-3 and setting y'=0 we get i and -i as the solutions. What does this say about the existence of the max and min points for the function y?

- Kamataat
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Popey
#2
Dec17-04, 11:26 AM
P: 22
Hi!

-3x2-3=0 has no solutions in real numbers

So, y' is always negative (as -3 is)
Hence, y is always decreasing (no min and max)
Kamataat
#3
Dec18-04, 04:31 AM
P: 137
ok, thanks

- Kamataat

tongos
#4
Dec18-04, 09:42 PM
P: 84
Maxima-minima problems

-x^3-3x=k

-x^3-3x-k=0=> where b makes x only have two solutions.....
x^3+3x-3x^2+k=(x^2-bx-(b/2)^2)(x+c)
x^3-bx^2-b^4/4x+x^2c-bcx-cb^4/4
k= -cb^4/4
-b+c=0
-b^4/4-b^2=-3
b^4+4b^2=12
b^4+4b^2-12=0
(b^2-h)(b^2-a)
(a+h)=-4
ah=12
crisalyn
#5
Sep17-10, 03:16 AM
P: 1
hi!
how determine whether that point is the maximum or the minimum?
Mark44
#6
Sep17-10, 09:10 AM
Mentor
P: 21,214
Quote Quote by crisalyn View Post
hi!
how determine whether that point is the maximum or the minimum?
The second derivative test is helpful. At a critical number c for which f'(c) = 0, if f''(c) > 0, (c, f(c)) is a local minimum point; if f''(c) < 0, (c, f(c)) is a local maximum point.

There's more to this, but your calculus text should have more information about the details.

In the future, if you have a question, start a new thread rather than adding onto an old thread. This thread is six years old.


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