# Very simple question

1. May 5, 2005

### semidevil

i'm trying to do analysis, but the stupid case by case part is confusing me and it's keeping me away from doing calculus.

ex:
h(x) = 3x^2 - 3
h'(x) = 3(x+1)(x-1)

how did they break down h'x to analyze it?

like, for x <= -1, --> h'x >= 0 --> h is increasing
-1<x<1 --> h'x <= 0 --> h is decreasing..

etc etc....

stuff like that...how did they get h'x is greater or lesser then 0 and increasing decreasing?

also

|x| + |x+1|

how did they get x < -1 implies |x| = -x and |x + 1| = -x - 1?

i'm sure it is really simple, but i can't see it.

2. May 5, 2005

If $h(x) = 3x^2 - 3 [/tex] $$h'(x) = 6x$$ Critical point is at x = 0 Decreasing at x < 0, Increasing at x > 0. 3. May 5, 2005 ### Data What does [itex]|x|$ mean?

As whozum noted, your derivative is wrong for the first question.

4. May 5, 2005

### Jameson

There has to be some typo, if $$h(x) = 3x^2-3$$, then $$h'(x)=6x$$
The $$3(x+1)(x-1)$$ is just a factored form of h(x).

EDIT: It seems the other posters already cleared up this question. Fix your derivative.

5. May 5, 2005

### semidevil

sorry, I meant h'x = 3x^2 - 3.

and factor that to get 3(x+1)(x-1)...how did all the cases come up?

6. May 6, 2005

### Data

well, when is x-1 positive? When is x+1 positive? From that, can you tell when their product is positive?

7. May 6, 2005

### dextercioby

Or simply plot that parabola.You know where its vertex is and you can find its intercepts with the Ox axis...

Daniel.