# Continuity (real analysis)

1. Mar 26, 2008

### Necrologist

1. The problem statement, all variables and given/known data

f(x) = 4 for x > or = 0, f(x) = 0 for x < 0, and g(x) = x^2 for all x.
Thus dom(f) = dom(g) = R.

2. Relevant equations

a. Determine the following functions: f+g, fg, f o g, g o f. Be sure to specify thier domain.

b. Which of the functions f, g, f+g, fg, f o g, g o f is continuous

3. The attempt at a solution

Ok, so for part (a) I am at f o g step, which I say is f(x^2) = 4 for x > or = 0 and f(x^2) = 0 for x < 0.

Question 1. Since f is a function of g(x), then when I restrict it values, do I have to restrict them in terms of x (as given) or in terms of x^2, since it is a f of g(x) now? Would I have to say that f(x^2) = 4 for x^2 > or = 0 and f(x^2) = 0 for x^2 < 0, instead of how I provided above? If I do have to say it in terms of x^2, then x^2 < 0 makes no sense, and the f would not be defined as f(g(x)) < 0?

Question 2. Since the composition f o g from Q1 is not piecewise anymore and starts at 0 and goes to positive infinity, would it make the composition continuous function?

Question 3. I am a little confused on how to prove that f is not a continuous function using limits. Would someone be able to show me a quick example please?

Question 4. If f is not continuous, then I assume that f, g, f+g, fg are not continuous either. But if f o g IS continuous (Q3), then wouldnt it contradict a Theorem that states "If f is continuous at x0 and g is continuous at f(x0), then the composite functions g o f is continuous at x0"?

2. Mar 27, 2008

### HallsofIvy

Staff Emeritus
Did you check that at all? Suppose x= -1. What you have written says that f(1)= 0. Any any case, what you have given looks like a definition of f. You want to write f o g(x)= something. In this case, that is particularly easy, because g(x) is never negative: f o g(x)= f(g(x))= f(x^2)= 4 for all x. Now, if x$\ge$0, what is g o f(x)= g(f(x))? If x< 0, what is g o f(x)?

No, f is not a function of g(x)! f is defined as a function of x. And, in fact, so are f o g and g o f. But don't confuse those.

No, "it", if you mean f itself, is not "f of g(x)".

There is NO f(x^2), only f(x). Again, don't confuse f o g with f. In fact, f o g(x)= 4 for all x.

f o g is (piecewise) continuous and does NOT "start at 0 and go to positive infinity". f o g is the constant function f o g(x)= 4. That's trivially continuous.

Suppose f(x)= 1 if x< 0, 2 if x$\ge$0. Then $\lim_{x\rightarrow 0^-}f(x)= 1$ while $\lim_{x\rightarrow 0^+}f(x)= 2$. Since those are not the same, $\lim_{x\rightarrow 0} f(x)$ does not exist so we do not have $lim_{x\rightarrow 0} f(x)= f(0)$. f(x) is not continuous at x= 0.

I recommend you not assume that!
No, the theorem says what happens what happens if f and g are continuous- it says nothing about what can happen if either is not continuous.

3. Mar 27, 2008

### Necrologist

so for g o f(x) do we have g(f(x))=
g(4)=x^2 for x > or = 0 and g(0)=x^2 for x < 0 ?

And dom(g o f) = {0, 4}?

The graph would be only two points?

4. Mar 28, 2008

### HallsofIvy

Staff Emeritus
Does g(4)= x^2 and g(0)= x^2 make any sense at all? g(4) is a NUMBER, not a variable. You were told that g(x)= x2 so g(4)= 42= 16.

For any x, g o f(x) means g(f(x)). Now your definition of f says that "f(x) = 4 for x > or = 0, f(x) = 0 for x < 0" so that if x>= 0, g(f(x))= g(4)= 42= 16. If x< 0, f(x)= 0 so g(f(x))= g(0)= 02= 0.

g o f(x)= 16 if x>= 0, 0 if x< 0. It's domain is all real numbers.