# Thread: Concave, Inflection Points, Etc

1. ## Concave, Inflection Points, Etc

Here's the problem:
[img][/ihttp://img641.imageshack.us/img641/645/screenshot20100321at540.pngmg]

a. ON what interval(s) is the function concave up?
b. ON what interval(s) is the function concave down?
c. FIND all points of inflection (if any).

So, the first thing I would do is to find the first derivative by doing the whole quotient-rule thing, which gives me:

The next thing I [think I] have to do is to take the derivative of that (second derivative). After doing the quotient rule of that, I get:

I'm sort of lost as to what I need to do next. I've been on spring break for a week and have forgotten a lot of this stuff... Looking back at previous problems just seem to confuse me.

2. Originally Posted by BeSweeet
Here's the problem:

a. ON what interval(s) is the function concave up?
b. ON what interval(s) is the function concave down?
c. FIND all points of inflection (if any).

So, the first thing I would do is to find the first derivative by doing the whole quotient-rule thing, which gives me:

The next thing I [think I] have to do is to take the derivative of that (second derivative). After doing the quotient rule of that, I get:

I'm sort of lost as to what I need to do next. I've been on spring break for a week and have forgotten a lot of this stuff... Looking back at previous problems just seem to confuse me.
set g''(x) = 0 ... inflection points on the graph of g(x) occur where g''(x) changes sign.

3. So it would be +$\displaystyle sqrt(3)$ and -$\displaystyle sqrt(3)$?

4. Originally Posted by BeSweeet
So it would be +$\displaystyle sqrt(3)$ and -$\displaystyle sqrt(3)$?
no ... $\displaystyle x^2+3 = 0$ has no real solutions

5. How do you know that? Sorry for the extremely noob questions...

So there aren't any inflection points? What about the concavity?

6. Originally Posted by BeSweeet
How do you know that? Sorry for the extremely noob questions...

So there aren't any inflection points? What about the concavity?
I did not say that ... I just told you that your algebra is wrong.

7. Hmm...

So to do this, I set the second derivative to 0, multiply the denominator by both sides, which leaves me with the numerator $\displaystyle 6x(x^2+3)$. I get 0, and +/-sqrt(3), which I guess isn't real, so I just have 0... I'm really not getting any of this stuff, nor do I know what any of this is for.

8. Originally Posted by BeSweeet
Hmm...

So to do this, I set the second derivative to 0, multiply the denominator by both sides, which leaves me with the numerator $\displaystyle 6x(x^2+3)$. I get 0, and +/-sqrt(3).
$\displaystyle x = 0$ ... correct

$\displaystyle x = \pm \sqrt{3}$ is incorrect.

look and learn ...

$\displaystyle x^2 + 3 = 0$

$\displaystyle x^2 = -3$

now you tell me, what number can you put in for x, square, and get -3 ???

9. Nothing. So the point of inflection is at 0?