Piecewise Problem: h(x) = |x-2| + |x+5|

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Homework Statement


h(x)= l x-2 l + l x+5 l


Homework Equations





The Attempt at a Solution

 
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What is your question?
 
I assume he's going to post it in pieces. Hence the name of the thread :)
 
ha okay, well I'm just completely lost on how to go about making this a piecewise.
I've never encountered this before, and I've compared it to others.
Apparently something just isn't clicking with me.
any assistance?
 
You mean you want to describe this function piecewise? It's not very clear, since it IS described piecewise right now (one piece).

You probably want to do something like this:

I can describe the function g(x)=|x| by the following:
g(x) = x if x>=0
g(x) = -x if x<0

So you need to find where things inside the absolute value sign change signs, and describe h(x) piecewise between those points (hint: they change sign when they're zero)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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