#Not sure how the problem solving logic works here
left moth
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safe anchor
To check whether a piecewise function is continuous, you need to check for continuity on the boundary points of the pieces. So, here you need to ensure:
f(-1-) = f(-1+)
f(2-) = f(2+)
fading finch
And combining (1) (2) and (3) you find the answer
left moth
To check whether a piecewise function is continuous, you need to check for conti...
so what your doing here is checking if there are limits that exist?
I previously understood this but I'm forgetting for some reason
safe anchor
so what your doing here is checking if there are limits that exist?
Not just that they exist, but that they are equal.
Recall that a function f(x) is continuous at x = x0 if f(x0-) = f(x0) = f(x0+).
Here's a flowchart, just in case.
left moth
so what they're basically using is condition 2 (limit of f(x) exists) in order to solve for m and n, but what would theoretically happen if the limit did not exist?
also I'm starting to hate this question the more I look at it because of it's ambiguity
safe anchor
so what they're basically using is condition 2 (limit of f(x) exists) in order t...
In that case the function would have a jump discontinuity for any values of m and n.
left moth
didn't indicate that the function was continuous, but instead asked to make it continuous
safe anchor
also I'm starting to hate this question the more I look at it because of it's am...
Not sure what you mean. The question is clear to me.
left moth
wait a minute is that why they set both limits equal to eachother?
safe anchor
didn't indicate that the function was continuous, but instead asked to *make* it...
That's because the function is continuous only for some values of m and n, not for all.
left moth
in order to make it continuous?
safe anchor
Yes.
left moth
but then why'd they throw in a -5 in there
safe anchor
In the statement?
safe anchor
Because the function is only defined for x ≥ -5.
left moth
not sure i understand
safe anchor
The domain of the function is x ≥ -5.
So, it only makes sense to talk about continuity on that set, not a larger one, as the function isn't defined outside of it.
left moth
hm okay, so it basically serves no purpose
safe anchor
Well, continuity is looked at on a certain set.
For example, if we have f(x) = 1/x, then it's continuous on its domain, but it clearly isn't continuous on the whole line.
left moth
okay let me try breaking down the logic of this question then:
- You need to make a non continuous function continuous
- To do so, you must set the limits of the end points equal to one another
- After you set the limits of the end points equal to one another you will find "m" and "n" by taking the limits for both sides
- Subtracting "m" and "n" to isolate for one variable will serve as a substitution and an answer to make the function g(x) continuous
safe anchor
okay let me try breaking down the logic of this question then:
- You need to mak...
You need to set the one sided limits at each point equal to each other. The limits at different points are independent.
You get a system and solve it for m and n.
left moth
only the first two steps are the thinky partsw
now you wouldn't do this unless you wanted to make the function continuous because setting two limits equal to one another is basically saying "yes this function is continuous at this value"
I think i get it now
left moth
hm okay wait
can a continuous function contain a hole?
because that's what I'm seeing in both of these points picked
safe anchor
Find the values of m and n for which this function is continuous:
f(x) = ...
...m(x - 1)^2 - n + 10, x < 0
...1, x = 0
...2m(x + 1)^3 - n, x > 0
safe anchor
can a continuous function contain a hole?
No. It is, after all, continuous.