Why is this function piecewise continuous and not just continuous on [0,10]

[Integrate]

Once we factor and cancel for t-2 we should be left with a discontinuity at t=-2 which is outside our interval of [0,10]. Meaning the function should just be continuous. Not Piece wise continuous.


How does this solution explain that it is piecewise continuous? Just that it can be both?

 

[Dr.Peterson]

Once we factor and cancel for t-2 we should be left with a discontinuity at t=-2 which is outside our interval of [0,10]. Meaning the function should just be continuous. Not Piece wise continuous.

Just pay attention to which function they are asking about:

• The simplified function r(t) is continuous on [0, 10].
• But the given function f(t) is not defined at t=2, so it can't be continuous on the entire interval.
• And the new function in the last line, in which they have redefined f(2) (with some value other than 1/4) is not continuous at t=2 because that value is not equal to the limits. But it is continuous on each of two pieces.

Does that answer your question?

 

[Steven G]

t-2 = 0 when t=2, NOT when t=-2.

 

[Integrate]

Just pay attention to which function they are asking about:

• The simplified function r(t) is continuous on [0, 10].
• But the given function f(t) is not defined at t=2, so it can't be continuous on the entire interval.
• And the new function in the last line, in which they have redefined f(2) (with some value other than 1/4) is not continuous at t=2 because that value is not equal to the limits. But it is continuous on each of two pieces.

Does that answer your question?

This really really helps.

I never really thought of a function being different from its simplified version.


Although, I'm not sure what's really going on in the last line.

 

[Dr.Peterson]

I never really thought of a function being different from its simplified version.

A function is a pairing of each input value in a given set (the domain) with an output value; it is a different function if it has a different set of input values (domain).

In cases like this, simplifying changes the domain, so it is no longer the same function.

More particularly, simplifying in this case results in a function with different properties (continuous, where the original was not).

Although, I'm not sure what's really going on in the last line.

Here's the last line:

​

What they're doing here is a sort of unsimplifying -- messing with the original function by adding 2 into its domain, but with a value that makes it not continuous.

Have you tried graphing the functions yet?

Here is the graph of [imath]\displaystyle f(t)=\frac{t^2-3t+2}{t^2-4}[/imath]:

​

Here is [imath]\displaystyle r(t)=\frac{t-1}{t+2}[/imath]:

​

It's a different function because it's defined for t=2.

Now here's [imath]\displaystyle g(t)=\left\{\begin{matrix}\frac{t^2-3t+2}{t^2-4} & t\ne2\\ 1 & t=2\end{matrix}\right.[/imath]:

​

There, I chose to let their "a" be 1. Do you see how this, too, is a different function?

 

[Integrate]

A function is a pairing of each input value in a given set (the domain) with an output value; it is a different function if it has a different set of input values (domain).

In cases like this, simplifying changes the domain, so it is no longer the same function.

More particularly, simplifying in this case results in a function with different properties (continuous, where the original was not).


Here's the last line:

View attachment 38118​

What they're doing here is a sort of unsimplifying -- messing with the original function by adding 2 into its domain, but with a value that makes it not continuous.

Have you tried graphing the functions yet?

Here is the graph of [imath]\displaystyle f(t)=\frac{t^2-3t+2}{t^2-4}[/imath]:

View attachment 38119​

Here is [imath]\displaystyle r(t)=\frac{t-1}{t+2}[/imath]:

View attachment 38120​

It's a different function because it's defined for t=2.

Now here's [imath]\displaystyle g(t)=\left\{\begin{matrix}\frac{t^2-3t+2}{t^2-4} & t\ne2\\ 1 & t=2\end{matrix}\right.[/imath]:

View attachment 38121​

There, I chose to let their "a" be 1. Do you see how this, too, is a different function?

Absolutely fantastic, thank you.