Absolute vs. Relative Maxima and Minima

Neen87
Messages
8
Reaction score
0

Homework Statement



I do not have a specific homework problem, but could someone please clarify this for me?

QUESTION: When you have an absolute maxima (or minima), how can you tell if it is ALSO a relative maxima (or minima)?

I understand how to find absolute extrema on a closed interval, and how to find critical values of a function by setting f'(x) = 0 and when f'(x) is undefined.


Thank you!
Tina
 
Last edited by a moderator:
Physics news on Phys.org
Neen87 said:

Homework Statement



I do not have a specific homework problem, but could someone please clarify this for me?

QUESTION: When you have an absolute maxima (or minima), how can you tell if it is ALSO a relative maxima (or minima)?

I understand how to find absolute extrema on a closed interval, and how to find critical values of a function by setting f'(x) = 0 and when f'(x) is undefined.


Thank you!
Tina
The question is usually asked the other way around. IOW, If you have a relative maximum (or minimum), how can you tell if it is ALSO an absolute maximum (minimum)?

It's generally easier to find relative maxima or minima by finding the values for which the derivative is zero or undefined and checking endpoints of the domain. From these points, it's just a matter of finding the overall largest and smallest to get the absolute maximum and minimum.
 
Thanks for your response!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top