Discontinuous Functions and Their Combinations: A Challenge in Continuity

[mathkillsalot]

Homework Statement

Give an example of a function f and g such that both are discontinuous at some value c in the Reals and a.)the sum f+g is continuous at c b.)the product fg is continuous at c

Homework Equations

...

The Attempt at a Solution

I have no idea as to how in the world i could answer this

 

[╔(σ_σ)╝]

Take the problem step by step.

First, if a function is discontinuous at a point c then there is some sort of "jump" ( there are different types of discontinuities but this is the one we are concerned with).

You want a function such that f and g are discontinous at c but f+g is continuous at c ( if a function is continuous at c it must be continuous around c).

Hint: What if f and g both had jumps but the sum f+g makes that jump disappear? Consider a simple case of constant functions with jumps.

 

[mathkillsalot]

okay, I've tried...am I right in using trial and error for this though??

 

[╔(σ_σ)╝]

In a sense you are using trial and error to get a mental picture of what the function looks like. I would say you are constructing a function not just randomly guessing.

However, you only use trial and error to "guess" the function but you still have to prove that the sum or product is continuous using the definition of continuity.

What do you have so far ? Can you show me ?

Consider the function f(x) =1 everywhere except 0 and f(0)=-1. Clearly, f is not continuous at 0. Can you find a function g(of the sane form as f) such that f+g would be continuous everwhere ?

 

[mathkillsalot]

i used the piecewise function:
f(x) = x^2 + 2 ; x<0
x^1/2 + 1 ; x>=0

g(x) = -3+x ; x<0
x-2 ; x>=0

they should be continuous at f+g (if I didn't do anything wrong), but I'm still working on fg

 

[╔(σ_σ)╝]

i used the piecewise function:
f(x) = x^2 + 2 ; x<0
x^1/2 + 1 ; x>=0

g(x) = -3+x ; x<0
x-2 ; x>=0

they should be continuous at f+g (if I didn't do anything wrong), but I'm still working on fg

Yes, this works for f+g.

Try simple functions like constant functions etc. The more complicated f and g getcthe harder it becomes to verify that f+g and fg are continous. Remember,you still have to prove that this functions are continuous.

 

[mathkillsalot]

but wait, I'm sort of confused as to how I do this.

How do I verify that the limit exists if it's x=0 or x(is unequal to)0 in:
lim f(x) = f(0)
x -> 0

and, how do I add and multiply? Do I choose f(0) or f(x)??

 

[mathkillsalot]

i mean, the limit doesn't approach 0. it actually is(or isn't) 0

 

[╔(σ_σ)╝]

but wait, I'm sort of confused as to how I do this.

How do I verify that the limit exists if it's x=0 or x(is unequal to)0 in:
lim f(x) = f(0)
x -> 0

and, how do I add and multiply? Do I choose f(0) or f(x)??

I am not sure what you are asking. Can you explain further?

I still encourage you to investigate into the following function...
f(x)=1 when x=/=0,f(0)=-1

Try to get a similar function for g.

 

[mathkillsalot]

what I mean is, how can we evaluate the one-sided limits if it's x=0 or x=/=0??

i really can't make it continuous at f+g and fg at the same time, I am using
f(x)
= 1 ; x>=0
=-1 ; x<0
g(x) = |2|
= 2 ; x>=0
= -2 ; x<0

I just made the conditions either x>=0 or x<0 so that I could evaluate their one sided limits

any suggestions?? and sorry, I'm not really that good at this...thanks so much for helping

 

[SammyS]

i used the piecewise function:
f(x) = x^2 + 2 ; x<0
x^1/2 + 1 ; x>=0

g(x) = -3+x ; x<0
x-2 ; x>=0

they should be continuous at f+g (if I didn't do anything wrong), but I'm still working on fg

To get fg to be continuous at x=0, try modifying your functions so that one of them, say f, goes to zero as x→0⁻ (from the left) the other, say g, goes to zero as x→0⁺ (from the right).

Then see if that works for the product function, fg.

If that works, try modifying the result to make each jump the opposite amount that the other jumps.
Do this without changing lim (x→0⁻) f(x) & lim (x→0⁺) g(x).

 

[mathkillsalot]

To get fg to be continuous at x=0, try modifying your functions so that one of them, say f, goes to zero as x→0⁻ (from the left) the other, say g, goes to zero as x→0⁺ (from the right).

Then see if that works for the product function, fg.

If that works, try modifying the result to make each jump the opposite amount that the other jumps.
Do this without changing lim (x→0⁻) f(x) & lim (x→0⁺) g(x).

thank you now I just pray that my prof will accept this answer

 

[╔(σ_σ)╝]

When I gave my hints I had a simpler solution in mind.

Consider the functions f and gf= \left\{\begin{array}{l}<br /> <br /> 1 \quad \quad \quad \text{ if } \quad x \neq 0\\<br /> <br /> -1 \quad \quad \text{ if } \quad x =0<br /> <br /> \end{array}\right.

g= \left\{\begin{array}{l}<br /> <br /> -1 \quad \quad \text{ if } \quad x \neq 0\\<br /> <br /> 1\quad \quad \quad \text{ if } \quad x =0<br /> <br /> \end{array}\right.

f+g =0 ; it is clear that f+g is everywhere continuous.

f\cdot g = -1 ; it is clear that fg is everywhere continuous.

 

[mathkillsalot]

When I gave my hints I had a simpler solution in mind.

Consider the functions f and g


f= \left\{\begin{array}{l}<br /> <br /> 1 \quad \quad \quad \text{ if } \quad x \neq 0\\<br /> <br /> -1 \quad \quad \text{ if } \quad x =0<br /> <br /> \end{array}\right.

g= \left\{\begin{array}{l}<br /> <br /> -1 \text{ if } \quad x \neq 0\\<br /> <br /> 1 \text{ if } \quad x =0<br /> <br /> \end{array}\right.

f+g =0 ; it is clear that f+g is everywhere continuous.

f\cdot g = -1 ; it is clear that fg is everywhere continuous.

oh wow. good god that never crossed my mind... TT_TT
awesome answer to the problem
was so caught up with the form
lim f(x) = f(a)
x -> a

thankyou )))

 

[╔(σ_σ)╝]

Well I hinted at it in post 4 and 9.

 

[mathkillsalot]

yes yes, but using the solution you just presented really never crossed my mind and I am sorry about that. Could've saved me some time...

Lesson learned, next time I'll...experiment on everything )
thank you dude!(if you are a dude) you are awesome XDDD

 

[╔(σ_σ)╝]

yes yes, but using the solution you just presented really never crossed my mind and I am sorry about that. Could've saved me some time...

Lesson learned, next time I'll...experiment on everything )
thank you dude!(if you are a dude) you are awesome XDDD

Yes I am a guy.

It was my pleasure.:D