How to see in what points the function changes

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in this function
http://img72.imageshack.us/img72/7038/31777281zi5.gif
its impossible to find the extreme points
y'(x)=0
x*2^(x-1)=4

how to find the points for which the function changes in sign??
 
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transgalactic said:
in this function
http://img72.imageshack.us/img72/7038/31777281zi5.gif
its impossible to find the extreme points
y'(x)=0
x*2^(x-1)=4

how to find the points for which the function changes in sign??
From your notes, f(x) = 2x - 4x + 1
f'(x) is not equal to x*2^(x - 1) - 4. You are using the power rule for derivatives, but that applies only to xn.

After you find the correct derivative, set it to 0. You might need to settle for approximate values for where f'(x) = 0.

Based on the points you found, there has to be a minimum somewhere between x = -10 and x = 10.

BTW, for such a simple function, I personally would like it better if you included all of the information on one page, rather than forcing readers to open another Web page to see 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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