# Maxima-minima problems

1. Dec 17, 2004

### Kamataat

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

2. Dec 17, 2004

### Popey

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)

3. Dec 18, 2004

### Kamataat

ok, thanks

- Kamataat

4. Dec 18, 2004

### tongos

-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

Last edited: Dec 19, 2004
5. Sep 17, 2010

### crisalyn

hi!
how determine whether that point is the maximum or the minimum?

6. Sep 17, 2010

### Staff: Mentor

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.