Question

Wanna confirm if it's right or not...
it's about the continuity. If there're two functions, f and g.
If f and g are continuous, then, is f(x) + x g(x) continuous?

I think the solution's continuous, but I'm not sure how to explain it. Somebody help?? Cheers xx

Answer 1

hmmm i had to know this like a few months ago hmm let me try to remember

Answer 2

the solution is continous that is a forsure

Answer 3

yeah, but how to explain it regarding to the rules etc?

Answer 4

if someone could explain it well

Answer 5

@candyfloss94 .... the function is continous...unless the limits are same or thy lie within the limit

Answer 6

well its the sum rule of differentiation maybe

Answer 7

this's random example to proof continuity. If h(x) is cont (in this case h(x) = x), then we can prove it by basic rules.... But further question, do you think h(x) will always be cont for other examples??

Answer 8

ur thin is correctt @ candyfloss94

Answer 9

What I mean by basic rules:
"If f and g are continuous at a and c is a constant, then the following also cont,
f+g

fg
f-g
cf
f/g if g(a) doesn't equal to 0"

Answer 10

Further question for you and others, do you think h(x) will always be cont for other examples??

Answer 11

@candyfloss94 ......no it wont be continous....unless ther exit a solution to it

Answer 12

Let \(f\) and \(g\) be continuous functions. Then for every \(\epsilon>0\), there is a \(\delta>0\) such that \(|x-x_0|<\delta\) implies \(|f(x)-f(x_0)|<\epsilon/2\) whenever \(x_0\) is in the domain of \(f\). This is stated similarly for \(g\). Then \(|f(x)+g(x)-f(x_0)-g(x_0)|\leq|f(x)-f(x_0)|+|g(x)-g(x_0)|<\epsilon/2+\epsilon/2=\epsilon\). Hence, \((f+g)(x)\) is also continuous.

Answer 13

@nasavivek sorry could you tell a bit more what do u mean with the prev reply? thanks x
@pre-algebra interesting, thanks! But I hv another proof in my book as well