Absolute Extrema of 2x - (x-2) on [0,1], [-3,4]

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    Absolute Extrema
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{} these brackets are going to represent the absolute value lines
the problem states
find the absolute extrema of the given function on each individual interval:
f(x)= {2x} - {x-2}
a) [0,1]
b) [-3, 4]

I know I need the derivative of the equation but it does not really give a good derivative since it would be f'(x)= 2 - {1}
 
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well, first what is {2x} equal to when x is from [0,1], als owhat is the value of {x-2}, do the same thing in the other interval!
 
You could also break the function up into the intervals (-\infty,0), [0,2), and [2,\infty) and write f as a piecewise function. Then, you can find the derivative on each of those open intervals (remember that the derivative won't necessarily be defined at 0 and 2).
 
how am i suppose to do tat
 
portillj said:
how am i suppose to do tat

Do u know how a piecewise defined function looks like? Well, to do that in this case you need to follow both my hints and also PingPong's hints!
 
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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