Continuity of a Function at x=-3 - Proving c Value

[KevinFan]

(Mentor note: moved from another forum hence no template)

For what value of the constant c is the following function continuous at x = −3?
f(x)=(1/x+1/3)/(x+3) if x≠ -3
f(x)=c if x=-3please provide proof... I am so confused:(

 

[fresh_42]

For what value of the constant c is the following function continuous at x = −3?
f(x)=(1/x+1/3)/(x+3) if x≠ -3
f(x)=c if x=-3please provide proof... I am so confused:(

Please provide the reason of your confusion ...

 

[KevinFan]

Please provide the reason of your confusion ...

I am confused because the second part of the function is just "c" which is a constant does not involve x.

 

[fresh_42]

What happens with $f(x)$ for $x \neq -3$ if you simplify the fractions? Have you tried to draw a graph around $x = -3$?

 

[KevinFan]

What happens with $f(x)$ for $x \neq -3$ if you simplify the fractions? Have you tried to draw a graph around $x = -3$?

I have tried to simplfy the function, it is f(x)=1/(3x) and I noticed for the simplfied function, x can be -3

 

[KevinFan]

What happens with $f(x)$ for $x \neq -3$ if you simplify the fractions? Have you tried to draw a graph around $x = -3$?

oh, is C=1/(3(-3))=-1/9 ??

 

[fresh_42]

And what is $f(-3)=c$ in this version? By the way: $x=0$ is also forbidden in the original definition of $x$.

 

[fresh_42]

oh, is C=1/(3(-3))=-1/9 ??

Yes. So now you have to find an argument, why $f(x)$ becomes continuous if we set $f(-3)= -\frac{1}{9}$.

 

[KevinFan]

Yes. So now you have to find an argument, why $f(x)$ becomes continuous if we set $f(-3)= -\frac{1}{9}$.

Many thanks for your help !

 

[KevinFan]

Yes. So now you have to find an argument, why $f(x)$ becomes continuous if we set $f(-3)= -\frac{1}{9}$.

On the oringinal function when x= -3, the function is undefined. However, if we set f(-3)=-1/9 then the oringinal function will become continuous on x=-3.
I think I understand now, thank you again for your help