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

momogiri
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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!
 
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1. set your derivative equal to 0 and solve for x
 
momogiri said:
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!

rocophysics said:
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?
 
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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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