How Do You Find Absolute Extrema of a Polynomial Function?

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simple SIMPLE local extremia question

Homework Statement



Find the absolute maximum and absolute minimum values of f(x) = 3x^4 - 8x ^3 + 12x^2 -48x +25, where 0 <= x <= 3.


Homework Equations



N/a

The Attempt at a Solution



f'(x) = 12x^3 - 24x^2 + 24x -48
= 12 (x^3 - 2x^2 + 2x -4)
= 12 (x-2) (X^2 +2)

I know how to do the rest, but just wondering before I go on... What should i do with that one factor...do I omit the (x^2 +2)?
 
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Well beside your first three lines of solution being irrelevant, it looks correct to me.

Edit: Actually you forgot a factor of 2 in your final h solution according to the symbolic result you derived earlier on.
 
Okay thanks alot, I updated the thread with a different question lol
 
Well you know that the absolute maximum/minimum lies at a place where the derivative is 0 or at the endpoints of the interval. Does x^2+2=0 have a real solution?
 
haha...thanks man. I'm starting to forget to most simple stuff
 
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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