Finding Values of a Function (increasing, decreasing, IPs, etc.)

[momogiri]

So the function is $$f(x) = 2 + 3x^{2} - x^{4}$$

Find the intervals of increase + decrease, local max + min value, inflection points (IP), interval the function is concave up + down

I know that I need to first find f'(x) to find the increase and decrease, so I solved that:

$$f'(x) = 6x - 4x^{3}$$

Now, the problem I'm having is finding the critical numbers. I know that 0 is obviously one of them, but I don't know what the other one is (since I'm quite confident that there is another one)
Thanks for your help!

 

[rocomath]

1. set your derivative equal to 0 and solve for x

 

[HallsofIvy]

So the function is $$f(x) = 2 + 3x^{2} - x^{4}$$

Find the intervals of increase + decrease, local max + min value, inflection points (IP), interval the function is concave up + down

I know that I need to first find f'(x) to find the increase and decrease, so I solved that:

$$f'(x) = 6x - 4x^{3}$$

Now, the problem I'm having is finding the critical numbers. I know that 0 is obviously one of them, but I don't know what the other one is (since I'm quite confident that there is another one)
Thanks for your help!

1. set your derivative equal to 0 and solve for x

I think that's what he said he did!

Yes, momogiri, f'(x)= 6x- 4x³= x(6- 4x²)= 0 at the critical points. It is because of that "x" factor that x= 0 is one. The other two, of course, satisfy 6- 4x²= 0. Can you solve that?

 

[momogiri]

Oh gosh, I can't believe I didn't see that XD
I'm pretty sure I can do it, Thanks for the help :D