Max points, points of inflection

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Homework Statement



which best describes y=4x^3 - 3x^4
find max/min points and inflection points

The Attempt at a Solution



When I work this one out I get x<0 and 0<x<1 as my two local max points. However, the book says there is only ONE max point - why is this?

with the second derivative I do get two inflection points of 0 and 1/2 which I assume to be correct.

Thoughts??
 
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I think that you'll see where your mistakes are once you write down a clear definition of an inflection point, and of a local extremum. I'll go ahead and say this (because it's often a step that students forget): did you check that your critical points are in fact local max/min points? Likewise for inflection points.

Can you also post your solution?
 
fitz_calc said:

Homework Statement



which best describes y=4x^3 - 3x^4
find max/min points and inflection points

The Attempt at a Solution



When I work this one out I get x<0 and 0<x<1 as my two local max points. However, the book says there is only ONE max point - why is this?
That makes no sense at all. Max/min points are individual points, not sets of points. HOW did you "work this one out"? If you mean that you got x= 0 and x= 1 as your max/min points, it is true that the derivative is 0 at x= 0 and x= 1, but that is not enough to be a max or a min.

with the second derivative I do get two inflection points of 0 and 1/2 which I assume to be correct.
That is not at all what I get. What is the second derivative?
 
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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