let f={4x for x<1;4x^2 for 1<=x <=3; -x for x>3}
Note f is continuous for x<1.
Note f is continuous at 1 as Lim(4x) = Lim(4x^2) as x--> 1.
Note f is continuous on (-inf,3).
However is f continuous on (1,3]?
I reason no.
By definition of continuity
1) the limit has to exist at f(3). The limit of 3 does not exist as right and left limits do not equal (as 4(3)^2 does not equal -(3))
A colleague has answered with
Yes the f is continuous on (1,3]. f=4x^3 so f'=12x^2 which is defined at 3. Differentiability implies continuity.
Help resolve the conflict.
Thank you.
Note f is continuous for x<1.
Note f is continuous at 1 as Lim(4x) = Lim(4x^2) as x--> 1.
Note f is continuous on (-inf,3).
However is f continuous on (1,3]?
I reason no.
By definition of continuity
1) the limit has to exist at f(3). The limit of 3 does not exist as right and left limits do not equal (as 4(3)^2 does not equal -(3))
A colleague has answered with
Yes the f is continuous on (1,3]. f=4x^3 so f'=12x^2 which is defined at 3. Differentiability implies continuity.
Help resolve the conflict.
Thank you.