How Do You Find Absolute Extrema of a Polynomial Function?

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To find the absolute extrema of the polynomial function f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25 on the interval [0, 3], the first derivative is calculated as f'(x) = 12x^3 - 24x^2 + 24x - 48. The critical points are determined by setting the derivative to zero, leading to the factorization 12(x-2)(x^2+2). Since x^2 + 2 has no real solutions, the only critical point is x = 2. The absolute maximum and minimum values must be evaluated at this critical point and the endpoints of the interval, x = 0 and x = 3.
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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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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