Identifying Points of Inflection for a Polynomial Function

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Homework Statement



Find the points of inflection of the graph of the function. (If an answer does not exist, enter DNE.)

Homework Equations


f(x)=1/3x^4 + 4x^3


The Attempt at a Solution


Found f'(x)= x^2(4/3x+12)
Also found critical numbers to x=0 and x=-9

I don't know how to get point of inflection
 
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Points of inflection are where the function changes concavity. They are related to second derivatives, aren't they? Look it up.
 
For inflection points:
Find the second derivative of the function and set it equal to zero.

I found that that x=-6 and x=0.

Test some points right and left of the second derivative zeros (^) to confirm concave up or concave down. A positive value is concave up, vice versa :)
 
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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