Graphing an Absolute Value Function

[nerd]

How is the best way to graph the following function. Other than addition of ordinates, is there an easier (algeraic) way?

[kurrymuncher]

Well you could find the line of symmetry for each absolute value and then graph each equation  according to that region. whether it be greater than x=2 or less than x=-4. You break the modulus up. gather all the equations that apply  to x<2, add them and graph it. You would then do this to  all the regions  (x>-4,  -4<x<2, x>2

This probably doesnt make sense.


I think someone like Mao can explain better, ask him.

[onlyfknhuman]

never seen this type of question before, i can break it up and and i found 2 equation. However when i check on the calculator im missing this big slope that connects it both no idea... algebraicly

[Tea.bag]

addition of ordinates.

[trinon]

There isn't a easier way to do it. Do it by addition of ordinates.. what's so hard about that?

[nerd]

It's ok...I found an algebraic way to do it! Thanks everyone for your help, though!

[Lulu]

It's ok...I found an algebraic way to do it! Thanks everyone for your help, though!

Want to explain how you did it so we can learn too?

[onlyfknhuman]

It's ok...I found an algebraic way to do it! Thanks everyone for your help, though!

Want to explain how you did it so we can learn too?

Yeah where too stupid to understand with one word

[bec]

What about finding the derivative of f(x)=|sin(2x)|?



Or do the domains need to be in terms of x?

[Mao]

firstly, if we look at the graph, f is not differentiable when sin(2x)=0 [it changes direction sharply], so it should not be

secondly, looking at your restraints, sin(x) is not necessarily applicable, but rather sin(2x).

I would agree with this set-up, though if you really wanted to change the restraints to in terms of x, you can:

, where

[Lulu]

I'm absolutely screwed for absolute functions!! (ha ha..see what I did there?)

[bec]

firstly, if we look at the graph, f is not differentiable when sin(2x)=0 [it changes direction sharply], so it should not be

secondly, looking at your restraints, sin(x) is not necessarily applicable, but rather sin(2x).

I would agree with this set-up, though if you really wanted to change the restraints to in terms of x, you can:

, where

So basically you're saying that, even though I got it all wrong, it's pretty much right?
haha i did mean <> and sin(2x) though, I just typed out both of them wrong...honestly...

[Chocoholic]

I'm absolutely screwed for absolute functions!! (ha ha..see what I did there?)

lol...you should absolutely go over absolute functions before th exam and definately revise definite integrals.

[nerd]

Actually, Mao, i think you missread the question.

It asks for



You have to look at the intercepts of the two absolute value functions and see where they intersect.

[danieltennis]

Isn't it not