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How do I find the maximum and minimum values of f(x)=x^3-5x^2+3x+12

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set the second derivative equal to zero.
Find the derivative of f(x) and set it equal to zero. Solve for x.
I thought it was the first derivative?

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Other answers:

yeah, you're right. the second derivative is to find if it is a min or a max.
Good point :D
So how exactly do I find the first derivative?
have you had calculus?
First of you know what a derivative is right?
No, im in algebra 2
The way my teacher taught how to find the min and max is by using a calculator , but if there is an easier way by hand, I wanna know howw
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this function is in odd degree so it has no max or min
How do I find the min and max using a calculator then?
if the function was in even degree has a max or a min or both like that i draw|dw:1358815524977:dw|
instead of f(x) use y. also what kind of calculator do you have?
so what should I do?
minimum: min{f(x) = x^3-5 x^2+3 x+12} = 3 at x = 3 max: max{f(x) = x^3-5 x^2+3 x+12} = 337/27 at x = 1/3

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