## anonymous one year ago Math help!

1. ybarrap

Do you know calculus? Or do you need to graph to solve?

2. anonymous

I know calculus. I have to get the second derivative of this equation correct? And then set it equal to 0 and find the x-values?

3. ybarrap

If x is an inflection point for y then the second derivative, y″(x), is equal to zero if it exists, but this condition does not provide a sufficient definition of a point of inflection. That's to say that if it y″(x) = 0, then it is not necessarily an inflection point. You are looking for the point where the function goes from concave up to concave down or vice versa. See red in this graph - https://en.wikipedia.org/wiki/Inflection_point#/media/File:Animated_illustration_of_inflection_point.gif

4. ybarrap

Here is a graph of your function. You can see where your function is concave up and concave down. Where it switches (from down to up or vice versa) is where you will test that in fact the second derivative is zero at that point and confirm that it is an inflection point.

5. anonymous

Wheres the graph of the function?

6. anonymous

It concaves up at 3

7. ybarrap

|dw:1436129525084:dw|

8. ybarrap

Where it goes from Concave down to Concave up is your inflection point

9. ybarrap

Now take the second derivative and set equal to zero. The point in that vicinity is the inflection point.

10. anonymous

I got -3 and 0

11. ybarrap

The second derivative set equal to zero is d^2/dx^2(x^3+18 x^2-3 x+4) = 6 (x+6) =0 So x = -6, right?

12. ybarrap

A little cleaner -- $$\frac{d^2}{dx^2}\left (x^3+18 x^2-3 x+4 \right )=6(x+6)=0$$ This is your equation, right?

13. ybarrap

Here is the second derivative - http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2+%28x%5E3+%2B+18x%5E2+%E2%80%93+3x+%2B+4%29

14. ybarrap

Does this make sense?

15. anonymous

Okay yeah I messed up when deriving

16. anonymous

Thank you!

17. ybarrap

You're welcome