Proving Continuity of Absolute Value Functions | Adv. Calc 1 Homework

[chief12]

Homework Statement

part 1)Show the function a(x)=|x| is a continuous function from R to R;

part 2)

Prove that if the functions f: D--> R is continuous at x=a, then l f l (absolute value of f) is also continuous at x=a.

Homework Equations

The Attempt at a Solution

part 1) since a(x) = |x|, then given any $$\epsilon$$ > 0, then for all l f(x) - f(a) l < $$\epsilon$$, since l x - a l < $$\delta$$ when delta = epsilon, since f(x) = l x l


part 2) since f is continuous, then the absolute value of f is also continuous since it doesn't change any of the relationships

then given any $$\epsilon$$ > 0, then for all l f(x) - f(a) l < $$\epsilon$$, since l x - a l < $$\delta$$ when delta = epsilon, since f(x) = l x l

 

[LCKurtz]

Homework Statement

part 1)Show the function a(x)=|x| is a continuous function from R to R;

part 2)

Prove that if the functions f: D--> R is continuous at x=a, then l f l (absolute value of f) is also continuous at x=a.

Homework Equations

The Attempt at a Solution

part 1) since a(x) = |x|, then given any $$\epsilon$$ > 0, then for all l f(x) - f(a) l < $$\epsilon$$, since l x - a l < $$\delta$$ when delta = epsilon, since f(x) = l x l

Not quite. Both the argument and the writeup need improvement.

Your final proof for part 1 should look something like this:

Suppose ε > 0. Pick δ = ε (if that is what works). Then if |x-a| < δ you have... At this point you need to give an argument to show that

|f(x) - f(a)| = | |x| - |a| |< ε

I used f(x) instead of a(x) for your function so it doesn't use "a" twice.