Graphing: test 2x^3 + 6x for concavity, infl. points

[andreipanait]

The question states:

Test f(x)= 2x^3 + 6x for concavity and inflection points. Show your work.

In order to test for concavity we first need to take the derivative, find the roots and then find the second derivative and then plug the roots into the 2nd derivative and see if it's positive or negative.

In this question, the problem is that, once I solve for the first derivative, I get 6x^2 + 6, which I then get 6(x^2 + 1). I'm not sure how to find the roots now and how to find the inflection points

~Thanks for your help

 

[stapel]

I'm not sure what you mean by "solving for the first derivative". Do you mean "finding (taking) the first derivative", or "solving for the zeroes of the first derivative", or something else? Also, what do you mean when you refer to plugging the zeroes of the second derivative back into the second derivative. Since this must, by definition, return a value of zero, what are you trying to accomplish by this?

Regarding the exercise: If the first derivative never changes signs, what does this tell you about the graph? What did you get for the second derivative? From this, what did you get for answers to the exercise (that is, what did you get for the inflection point(s) and concavity)?

When you reply, please clearly show your work and reasoning. Thank you.

Eliz.

 

[galactus]

f'(x) is positive \(\displaystyle \forall{x} \;\ \in \;\ \Re\)

Check the second derivative. Where does it equal 0?. That will be the points of inflection provided concavity changes there.

 

[andreipanait]

After reading what I wrote I realize that it doesnt make much sense, so I'll re-phrase it.

The original equation is 2x^3 + 6x

I took the first and then the 2nd derivative of this equation:

FIRST DERIVATIVE: 6x^2 + 6

SECOND DERIVATIVE: 12x

Ok, this is where I get confizzled. The 2nd derivative states that in order to find if the concavity:

f(x) > 0 it's concave up and f(x) < 0 concave down

 

[andreipanait]

f'(x) is positive \(\displaystyle \forall{x} \;\ \in \;\ \Re\)

Check the second derivative. Where does it equal 0?. That will be the points of inflection provided concavity changes there.

well the 2nd derivative is f''(x)=12x
so it will be f''(x)=0 when x is 0. That doesnt help me much im still confused
:S

 

[andreipanait]

ok i get it now. Because the 2nd derivative is 12x, the infliction point is 0 and thereby its concave up to the right of the infliction point and cconcave down to the left. Thanks for ur help I know I'm a little bit slow
Hi steve!