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Mathematics 7 Online
OpenStudy (anonymous):

How do I find the absolute maximum and minimum of the function, if I am just given a function f(x), without the graph?

OpenStudy (anonymous):

(I know how to derive)

OpenStudy (johnweldon1993):

Ah my first question^ First step is to take the first derivative of the function...then set that equal to 0 This will give you "critical points" which are potential maxima and minima

OpenStudy (anonymous):

yes, I know (critical points are also those at which f(x) doesn't exist. )

OpenStudy (anonymous):

And how do I know which of the critical points is the abs max and min ?

OpenStudy (johnweldon1993):

Indeed^ Following this step we take the second derivative of the function The second derivative at 'x' will give you which of those points are maxima or minima (or a saddle point) If the second derivative turns out negative we have a minimum If it turns out negative, we have a maximum

OpenStudy (anonymous):

I don't really get it. Can I do an example and you correct me if anything. Is that alright?

OpenStudy (johnweldon1993):

No problem!

OpenStudy (anonymous):

Lets say I have f(x)=x^4+x^3-2x+1

OpenStudy (anonymous):

bad example

OpenStudy (anonymous):

I;ll get an example and do it in another question. I;ll get the book. sorry abt that

OpenStudy (anonymous):

tnx for the help

OpenStudy (johnweldon1993):

Anytime

OpenStudy (johnweldon1993):

But for reference, we can do that example x^4 + x^3 - 2x + 1 First derivative would be 4x^3 + 3x^2 - 2 Set that = to 0 4x^3 + 3x^2 - 2 = 0 4x^3 + 3x^2 = 2 x(4x^2 + 3x) = 2 I'll just skip the process of completing the square here and all that and I get x is roughly equal to 0.60701 That is our 1 and only critical point that I found here So we just take the second derivative now...and plug in That for 'x' second derivative would be 12x^2 + 6x Plugging in x = 0.60701 we get a positive number...thus...a minimum

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