Find Absolute Maxima and Minima of f(x) on [-2,2]

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1. Find the absolute maxima and the absolute minima of the following function

f(x)=(2x)/(x^2+1) on [-2,2]
 
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i found the derivative = (2x^2-2)/(x^2+1)^2
 
Okay, what will the derivative be at a maximum point?
 
I can't solve it
 
A maximum would occur if the derivative exists and changes sign from positive to negative at a certain point.

A minimum would occur if the derivative exists and changes sign from negative to positive at a certain point.

This means the derivative must pass through zero to change the sign. So, equate your derivative to zero and solve for the x coordinates of these points. In Calculus, these points are called critical points.

Just in case you don't understand, I'd recommend the following resources:

http://ltcconline.net/greenl/courses/115/applications/frsttst.htm
http://www.math.wvu.edu/~hjlai/Teaching/Tip-Pdf/Tip1-21.pdf
http://www.math.ucdavis.edu/~xiaoh/16a/extrema.pdf

It's important to understand what the assignment is on before attempting it.
 
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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...
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