Thanks for the replies! I fixed up the errors in my thread and added parentheses.
so I re-did the left for x<0
I got
f' = (-(x+h) + (x+h) - (-x+x))/h = (-x - h + x + h + x - x)/h = 2x/h but taking the limit as h->0 it is undefined
is it an algebraic error I am making?
also when I graph it out it only shows the right side / instead of V
I don't understand how I got the left derivative as -1 when there's nothing on the left
Homework Statement
f(x) = |x| + x
Does f'(0) exist? Does f'(x) exist for values of x other than 0?
This is from lang's a first course in calculus page 54 # 13
Homework Equations
lim (f(x+h) - f(x))/h
h->0
The Attempt at a Solution
So I'm not sure if I am doing this...
Would this be classified as a direct proof? I'm trying to learn proofs on my own so this is a little bit confusing to me. Thanks everyone for helping me out!
okay so here is what I'm doing right now
|x+y| >= |x| - |y|
with x = x + y - y
I got
|x + y| >= |x| + |y| - |y| - |y|
cancelling the |y|
|x + y| >= |x| + |y|
am I on the right track? :)
Thank you, I understand why x = x + y - y it is very obvious! I am confused because I have no idea what it has to do with the proof and where to apply it when proving that |x+y| >= |x| - |y|
I don't understand why I would write x + y ≥ x + y
how does the subtraction turn into addition? does this mean that x + y = |x| - |y|? I see that -(-x) = + x
but its |x| - |y| not |x| + |-y|
or is my thinking wrong?
Homework Statement
|x + y| ≥ |x| - |y| [Hint: write out x = x + y - y, and apply Theorem 3, together with the fact that |-y| = |y|]
Homework Equations
Theorem 3: |a + b| ≤ |a| + |b|
x = x + y - y
|-y| = |y|
The Attempt at a Solution
|x + y| ≥ |x| - |y|
x = x + y - y (don't know where to...
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