# Increasing/decreasing, minima/maxima

• Oct 12th 2009, 05:43 PM
nautica17
Increasing/decreasing, minima/maxima
Okay, so I'm trying to find whether the graph is increasing/decreasing, and I need the minima/maxima.

My original function was x^3 - 4x

First derivative: 3x^2 - 4

I've gotten down to sqrt(4/3). Now, from here I am confused on exactly how to get what I mentioned above (increasing, etc.) I can do it with other types of functions but I don't exactly know what to do in a situation like this. Any help, please?

And as a side question, can someone calculate the concavity and inflection points and see if they get (0,0) for inflection and (1,inf) for concave up and (-inf, 1) for concave down? I want to check if I am at least doing that right.
• Oct 12th 2009, 05:49 PM
Arturo_026
it's +sqrt(4/3) and -sqrt(4/3)

then you set your intervals and get the second derivative to determine where on the intervals is the function increasing/decresing and what kind of concavity f has on each interval
• Oct 12th 2009, 05:53 PM
apcalculus
Quote:

Originally Posted by nautica17
Okay, so I'm trying to find whether the graph is increasing/decreasing, and I need the minima/maxima.

My original function was x^3 - 4x

First derivative: 3x^2 - 4

I've gotten down to sqrt(4/3). Now, from here I am confused on exactly how to get what I mentioned above (increasing, etc.) I can do it with other types of functions but I don't exactly know what to do in a situation like this. Any help, please?

And as a side question, can someone calculate the concavity and inflection points and see if they get (0,0) for inflection and (1,inf) for concave up and (-inf, 1) for concave down? I want to check if I am at least doing that right.

The first derivative has two zeros: $\displaystyle \pm \sqrt{\frac{4}{3}}$, so you'll need to study the sign of the first derivative in the intervals from negative infinity to $\displaystyle -\sqrt{\frac{4}{3}}$, between $\displaystyle -\sqrt{\frac{4}{3}}$ and $\displaystyle \sqrt{\frac{4}{3}}$, and from $\displaystyle \sqrt{\frac{4}{3}}$ to positive infinity. The original function will be increasing when the first derivative is positive, and decreasing when negative.

As far as inflection points go, these are points where the second derivative changes sign while the first derivative maintains its sign.

Good luck!
• Oct 12th 2009, 05:53 PM
nautica17
Okay, I figured it would something like that. Now how would I put that in notation?

Is this correct?

(-inf, -sqrt(4/3)) , (+sqrt(4/3),inf)
• Oct 12th 2009, 05:55 PM
Arturo_026