Maxima and Minima of a function


by n0_3sc
Tags: function, maxima, minima
n0_3sc
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#1
Jan9-09, 11:26 PM
P: 265
Are there any analytical techniques to do this besides the Derivative Test?
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HallsofIvy
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#2
Jan10-09, 06:34 AM
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There is no general method except by checking where the derivitive is 0 (or does not exist). For some functions, there are other ways. For example we can always find minima and maxima for quadratic functions by completing the square.
n0_3sc
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#3
Jan10-09, 12:27 PM
P: 265
I see. Thanks for that.

lukaszh
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#4
Jan11-09, 07:53 AM
P: 32

Maxima and Minima of a function


But there is also possibility to estimate. If you solve some elementary function, for example:
[tex]f(x)=x^2+3x+2[/tex]
You can transform it to form:
[tex]f(x)+\frac{1}{4}=\left(x+\frac{3}{2}\right)^2[/tex]
So now you are able to find a minimum:
[tex]\min_{x\in\mathbb{R}}f(x)=-\frac{1}{4}[/tex]
n0_3sc
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#5
Jan11-09, 03:05 PM
P: 265
Yes, but my function is far too complex/tedious to do either way. An expression for the min and max has been found though proving it is too difficult for me.
NoMoreExams
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#6
Jan11-09, 03:21 PM
P: 626
Quote Quote by lukaszh View Post
But there is also possibility to estimate. If you solve some elementary function, for example:
[tex]f(x)=x^2+3x+2[/tex]
You can transform it to form:
[tex]f(x)+\frac{1}{4}=\left(x+\frac{3}{2}\right)^2[/tex]
So now you are able to find a minimum:
[tex]\min_{x\in\mathbb{R}}f(x)=-\frac{1}{4}[/tex]
Well if you are working with quadratics a lot you should know that the min/max is going to be at [tex] \left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)[/tex]


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