Maxima and Minima of a function

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 5K views
n0_3sc
Messages
238
Reaction score
1
Are there any analytical techniques to do this besides the Derivative Test?
 
Physics news on Phys.org
There is no general method except by checking where the derivative 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.
 
I see. Thanks for that.
 
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]
 
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.
 
lukaszh said:
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]