Proving Continuous Function Homework

[lovemake1]

Homework Statement

Prove or give a counterexample for each of the following statements.

a) If f(x) is continuous, then the function lf(x)l is continuous.

b) if lf(x)l is continuous, then f(x) is continuous

Homework Equations

Given epsilon > 0 there is delta > 0 such that l x - c l < delta implies l f(x) - f(c) l < epsilon

The Attempt at a Solution

so we have to use the definition of continuity. ( lim x-> c f(x) = f(c) ) inaddition to
the delta epsilon method.

E = epsilon
S = delta
a)
lf(x) - cl < S implies lf(x) - f(c)l < E

lf(x) - f(c)l < E

f(c) - E < f(x) < f(c) + E

And since f(x) is continuos, we know that f(c) - E < lf(x)l < f(c) + E
will not change anything to the overall equation.

b) l lf(x)l - c l < S implies l lf(x)l - f(c) l < E

f(c) - E < lf(x)l < f(c) + E
this is continous, but if f(x) were -f(x)

f(c) - E < -f(x) < f(c) + E
this would change

-f(c) + E < f(x) < -f(c) - E

which is false, since -f(c) + E is bigger than -f(c) - E

is this right? please help !

 

[micromass]

Hmm, it seems you've got the right ideas. But I can't happen to follow your proof.

For 1. You need to find a $$\delta$$ such that

$$|x-c|<\delta~\Rightarrow~||f(x)|-|f(c)||<\epsilon$$

It is true that $$||f(x)|-|f(c)||\leq |f(x)-f(c)|$$ (prove this!). Since f is continuous we can find a delta such that $$|f(x)-f(c)|$$ is smaller then epsilon. But the inequality then yields that $$||f(x)|-|f(c)||\leq \epsilon$$ and that is what you had to show.

For 2. You are correct that this statement is false. But you still need to come up with a counterexample. Try finding a discontinuous function f such that |f(x)|=1 for all x. (There are loads of other counterexamples, so it doesn't need to be this one).