Continuity at Endpoints: let f={4x for x<1;4x^2 for 1<=x <=3; -x for x>3}

[datadan]

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.

 

[stapel]

\(\displaystyle \displaystyle f(x)\, =\, \begin{cases}4x&\mbox{ for}\, x\, <\, 1\\4x^2&\mbox{ for}\, 1\, \leq\, x\, \leq\, 3\\-x&\mbox{ for}\, x\, >\, 3\end{cases}\)

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))

I agree. The graph makes very clear that the second and third parts of the function do not join up, so the function is discontinuous at their break point.

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.

But doesn't the definition of the derivative at a point involve both right- and left-hand limits, which must be equal? But the limit, as x approaches 3 from the left, is y = 32; the limit from the right is y = -3. Since the two-sided limit does not exist, is the limit (and thus the derivative) well-defined?

 

[Dr.Peterson]

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.

The function f as defined is not continuous at 3, so it would be wrong to say that it is continuous on (1,3].

Your colleague is answering a slightly different question: The function f, restricted to the domain (1,3], is continuous on its domain. That's possible because this function is really defined as g(x) = {4x^2 for 1<x<=3}, which says nothing about values beyond 3, so that part is irrelevant. At an endpoint of a domain, one-sided continuity (which is all you can talk about) is sufficient. (The same is true of its being differentiable at 3.)

You stated part of the definition differently in the places I bolded; that doesn't directly affect my answer, but could be important for other reasons.