Finding out if piecewise function is 1) continous and 2) differentiable: method

[DeadmansswitchAAD]

Hi,

I'm wondering how to figure out if this piecewise function is continuous and differentiable.

"Is the function

1) continuous
2) differentiable

for all real numbers? Explain how you reached your conclusions"

Work I've done:
1) I set x=1 and get that the answer for both pieces is 1, and thus concluded that the function is continuous.
2) This is where I run into problems. I differentiated both pieces using the chain rule, and get that both pieces are equal to 1+(1/0), 1+infinity. What does this imply? Until now I've written that the derivative is undefined at x=1. However, the function is differentiable elsewhere. How do I go about answering 2)?

Are my conclusions correct?

From the given function I also know that the function overlap in this case, and thus the function is perfectly vertical here. Would it be correct to say that there is a "sharp" point here, and that the function is not differentiable as a result? If yes, is this the case for all continuous piecewise functions at x=a where there is an overlap between the two pieces?

 

[Dr.Peterson]

Hi,

I'm wondering how to figure out if this piecewise function is continuous and differentiable.

"Is the function

1) continuous
2) differentiable

for all real numbers? Explain how you reached your conclusions"

Work I've done:
1) I set x=1 and get that the answer for both pieces is 1, and thus concluded that the function is continuous.
2) This is where I run into problems. I differentiated both pieces using the chain rule, and get that both pieces are equal to 1+(1/0), 1+infinity. What does this imply? Until now I've written that the derivative is undefined at x=1. However, the function is differentiable elsewhere. How do I go about answering 2)?

Are my conclusions correct?

From the given function I also know that the function overlap in this case, and thus the function is perfectly vertical here. Would it be correct to say that there is a "sharp" point here, and that the function is not differentiable as a result? If yes, is this the case for all continuous piecewise functions at x=a where there is an overlap between the two pieces?

First, although your conclusions are correct, you should really express your thinking in terms of limits, not just plugging in values.

Second, you have answered (2): the function is differentiable everywhere except at x=1. The tangent is indeed vertical there, but I wouldn't call it a "sharp point". Would you like to show us your graph?

But what do you mean by "overlap"?

 

[DeadmansswitchAAD]

First, although your conclusions are correct, you should really express your thinking in terms of limits, not just plugging in values.

Second, you have answered (2): the function is differentiable everywhere except at x=1. The tangent is indeed vertical there, but I wouldn't call it a "sharp point". Would you like to show us your graph?

But what do you mean by "overlap"?

Orange is the graph of f(x), the grey is the graph of f'(x).

I just realized that "overlap" isn't the right term as the pieces don't "overlap". What I meant is that the function f(x) is defined at x=2 even though they're the pieces aren't both defined at x=2 individually. I'm finding it difficult to phrase this in precise terms.

 

[Dr.Peterson]

Orange is the graph of f(x), the grey is the graph of f'(x).

So, when you said, 'there is a "sharp" point here', were you talking about the graph of f'? The graph of f has nothing I'd call "pointy" at all; and the graph of f' has a vertical asymptote, not what I would call a "cusp", which may be what you have in mind.

I just realized that "overlap" isn't the right term as the pieces don't "overlap". What I meant is that the function f(x) is defined at x=2 even though they're the pieces aren't both defined at x=2 individually. I'm finding it difficult to phrase this in precise terms.

There are two "pieces" of the domain, which do not intersect (the formal word for "overlap") -- otherwise it wouldn't be a function at all. But they do "meet" (which is what makes the function continuous). When you say x=2, do you mean x=1? There is nothing special about 2 here.

I still don't know what you are trying to say. Try again to say it, in many, imprecise words!

From the given function I also know that the function overlap in this case, and thus the function is perfectly vertical here. Would it be correct to say that there is a "sharp" point here, and that the function is not differentiable as a result? If yes, is this the case for all continuous piecewise functions at x=a where there is an overlap between the two pieces?

What makes the function vertical is just that the derivative is infinite (undefined). The one thing I find interesting about the function is that it comes about as close to being continuously differentiable as you can come while not being differentiable at a point; it's like the cube root function, which likewise is vertical at one point.

Every continuous piecewise function will "meet" as this one does, whether or not it is vertical there. But one that is differentiable will have the same finite slope on both sides.

 

[DeadmansswitchAAD]

So, when you said, 'there is a "sharp" point here', were you talking about the graph of f'? The graph of f has nothing I'd call "pointy" at all; and the graph of f' has a vertical asymptote, not what I would call a "cusp", which may be what you have in mind.


There are two "pieces" of the domain, which do not intersect (the formal word for "overlap") -- otherwise it wouldn't be a function at all. But they do "meet" (which is what makes the function continuous). When you say x=2, do you mean x=1? There is nothing special about 2 here.

I still don't know what you are trying to say. Try again to say it, in many, imprecise words!

What makes the function vertical is just that the derivative is infinite (undefined). The one thing I find interesting about the function is that it comes about as close to being continuously differentiable as you can come while not being differentiable at a point; it's like the cube root function, which likewise is vertical at one point.

Every continuous piecewise function will "meet" as this one does, whether or not it is vertical there. But one that is differentiable will have the same finite slope on both sides.

“When you say x=2, do you mean x=1? There is nothing special about 2 here.”

Yeah, I was meaning to write x=1, not x=2.

“But they do "meet" (which is what makes the function continuous)”

“Meet”, yes.

“The graph of f has nothing I'd call "pointy" at all; and the graph of f' has a vertical asymptote (…) What makes the function vertical is just that the derivative is infinite (undefined)”

Oh, alright. You answered my poorly worded question:

(My quote): “Would it be correct to say that there is a sharp point here, and that the function is not differentiable as a result?”

(Based on my interpretation of what you’ve written, I believe should’ve worded this as “Would it be correct to say that the function f is not differentiable at x=1 due to the fact that the function f is vertical at x=1?”)

Thanks for your answers!

 

[Steven G]

A function continuous at x=a iff:

1a)$\displaystyle \lim_{x\to\ a^-} f(x)\ exists$
1b)$\displaystyle \lim_{x\to\ a^+} f(x)\ exists$
2)$\displaystyle \lim_{x\to\ a^-} f(x)\ =\lim_{x\to\ a^+} f(x)\ =L$
3) f(a) = L

You must verify all those statements above. Verifying some, most, a few or none of the statements above will not work.

 

[Steven G]

Work I've done:
1) I set x=1 and get that the answer for both pieces is 1, and thus concluded that the function is continuous.

Sorry but you can't conclude continuity just by what you say above. You also need, in this case, that f(1)=1.
If f(x) was NOT defined at x=1, even though when you plug in x=1 into each part and get 1, then f(x) is not continuous at x=1.

A necessary condition for f(x) to be continuous at x=1 is for f(x) to be defined there!