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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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