If f & g are continuous for all real x, will the sum of both be continuous?

Question

I know that if f and g are functions which are continuous at the point x = a, then the sum of f + g is continuous at x = a. But if f & g are continuous for all real x, will the sum & division of both be continuous for all real x as well? I think no, but I don't know how to prove it >_<
1 more question. If f(x)=g(x), for x isn't equal to c and f(c) isn't equal to g(c), then f & g aren 't continuous...is this true?I think it's false, but i don't know what function should I use.

Help please!
Thank you very much!

Answer 1

If f and g are continuous for all real x, then
f + g, f - g and f*g are continuous for all real x
f / g is continuous for all real x where g(x) ≠ 0.

If you know these properties are true pointwise (i.e. continuity of f and g at x=a implies continuity of f+g, f-g, f*g, and f/g for g ≠ 0 at x=a) then it follows immediately that they are true for continuity for all x, since "f(x) is continuous for all real x" is just a shorthand way of saying "f(x) is continuous at x=a for every real number a".

You can prove the properties formally with an ε-δ proof, but it's easier to do it with limits if you have proved the corresponding properties for limits. I won't go into the hairy details unless you want them.

If f(x) = g(x) for x ≠ c, and f(c) ≠ g(c), then f and g can't both be continuous at c, but one of them can be.

It's pretty simple to show that they can't both be continuous at c. Suppose that they were both continuous at c. Then we know that
lim (x->c) f(x) exists, call it L
and also lim (x->c) g(x) exists, call it K.
Since f(x) = g(x) for all x ≠ c, we must have L = K.

(If this isn't clear to you, we can prove it formally with an ε-δ proof as follows: Suppose L ≠ K, and let ε = |L-K|/2. Since both limits exist, there is a δ1 such that 0 < |x-c| < δ1 => |f(x) - L| < ε, by definition of the limit; similarly there is a δ2 such that 0 < |x-c| < δ2 => |g(x) - K| < ε. Choose δ = min{δ1, δ2}. Then 0 < |x-c| < δ => |f(x) - L| < ε and |g(x) - K| < ε, and f(x) = g(x), so we have
|L-K| <= |L - f(x)| + |f(x) - K| (triangle inequality)
< ε + ε = |L-K|. So |L-K| < |L-K|, a contradiction. So we must have L = K.)

So lim (x->c) f(x) = lim (x->c) g(x) = L.
But since f and g are continuous at c, f(c) = lim (x->c) f(x) and g(c) = lim (x->c) g(x), so f(c) = g(c) = L. But this contradicts the assumption that f(c) ≠ g(c). So f(x) and g(x) cannot both be continuous at c.

It's easy to find examples where one of the functions is continuous. Just take any continuous function for f(x), and let g(x) be the same as f(x) except at one point. For example, let f(x) = 0 for all x, g(x) = {0, all x ≠ 1; g(1) = 1}. Then f(x) = g(x) for all x ≠ 1; f(x) is continuous but g(x) is not.

Answer 2

I know that if f and g are functions which are continuous at the point x = a, then the sum of f + g is continuous at x = a. But if f & g are continuous for all real x, will the sum & division of both be continuous for all real x as well

Yes, the sum will be continuos, it's from the property that you already stated, since it is continuos pointwise, it is continuos on all domain.
Because by definition, a function is continuos on its domain if it is continuos on any point of the domain.
For division f/g will be continuos on any point x except point x
where g(x)=0, because of course, f/g is not defined on those points.

If f(x)=g(x), for x isn't equal to c and f(c) isn't equal to g(c), then f & g aren 't continuous...is this true?

OK, let's define like that
f(x) = g(x), if x not equal c
f(c) not equal g(c)
Now let's discuss the continuity of f depending on g
If g is continuos then f is not necessarily continuos

Example 1: g(x) = x continuos
f(x) = x, if x not equal 0
f(0) = 1
Clearly f is not continuos on 0, therefore is not continuous on R.

Now, case 2, if g is not continuos then f could be continuos or not.

Example 2: g(x)= x, for x not 0
g(0)=1
f(x) = g(x), for x not 1
f(1) = 2

both are discontinuos, g in 0, f in 0 and 1
Example 3:

g(x) = x, x not 0
g(0) =1
f(x) = g(x), x not 0
f(0)=0

Actually you have that f(x)=x, and this is the same with example 1( by symmetry f can be reversed with g)
So g is not continuos on 0 and f is continuos on R.

Answer 3

Continuity ability there is not any wreck interior the function. Derivable ability there is not any discontinuity interior the derivitive function. the two cosx and cos^3 x are non-end of themselves, however the question is what occurs at their factors of intersection. they are equivalent at x = 0, pi/2, pi, 3pi/2, 2pi etc the place their values are a million, 0 , -a million, 0 , a million The derivitive purposes are -sinx 3cos^2 x * (-sinx) = -3*sinx*cos^2 x At x = 0, the two derivites = 0 at x = pi/2, -sinx = -a million, the different one = 0 So the derivitives are actually not non-end at pi/2. regardless of the actuality that the function has the comparable fee, there's a leap interior the derivitive. So f(x) is non-end, yet no longer derivable at those factors.

Answer 4

yes the sum and product will also be continuous.