How To Find A Removable Discontinuity
How do I decide where a function has a removable and a jump discontinuity?
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Homework Statement
I am given the following function, piecewise:
f(x) = (-x+b) (10<1)
3 if x=ane
(-12/(10-b))-i (x>1, ten=/b)
I am asked:
ane) For what value(south) of 'b' does 'f' have a removable discontinuity at 1?
ii) For what value(s) of 'b' does 'f' have a (finite) bound aperture at 1? Write your answer in interval notation.
Homework Equations
X
The Attempt at a Solution
Honestly, I accept to clue where to start; especially in regards to the removable discontinuity. I tried making the functions equal each other for the jump discontinuity but, that was way out every bit far as I can tell.
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If I could at least be given a starting point to go from, it would be appreciated. I'm non asking for the answers only, I'yard so lost and just staring at this thing isn't helping. Thanks,
RK
Answers and Replies
[tex]
\begin{marshal*}
\lim_{x \to a^+} f(x) & = \lim_{x \to a^-} f(x)\\
\text{ so } & \lim_{x \to a} f(x) \text{ exists}\\
f(a) & \text{ is divers} \\
f(a) & \ne \lim_{x \to a^+} f(x)
\stop{align*}
[/tex]
All the limits in a higher place are finite. Basically, if there is a removable discontinuity at [tex] x = a [/tex], the part has almost everything needed to be continuous in that location, simply the function value is "incorrect". Here is an example.
[tex]
f(ten) = \begin{cases}
3x+five & \text{ if } 10 \ne 10\\
twenty & \text{ if } x = x
\end{cases}
[/tex]
Hither the limit at 10 is equal to 35 (because both one-sided limits equal 35), but [tex] f(ten) = xx [/tex], so the function is not continuous in that location. We say in that location is a removable discontinuity at 10 considering we tin can practice this: brand [tex] f [/tex] continuous just past doing this:
[tex]
F(x) = \brainstorm{cases}
3x + 5 & \text{ if } x \ne 10\\
35 & \text{ if } x = ten
\stop{cases}
[/tex]
i.e. - we remove the discontinuity by redefining the role at the problem point, in guild to make the function continuous there.
A function has a jump aperture at a spot if the two one-sided limits both exist but are not equal. Here is an example - the jump discontinuity is at [tex] ten = 2 [/tex].
[tex]
west(x) = \begin{cases}
2x - i & \text{ if } x <=ii\\
10x + ane & \text{ if } ten > two
\end{cases}
[/tex]
The limit from the left is three, the limit from the correct is 21. If y'all were to view the graph of [tex] west [/tex] you lot would see a 'jump' in the two portions of the graph at [tex] x = 2 [/tex]. What yous need to do is find the values of [tex] a [/tex] and [tex] b [/tex] to make your function take the types of beliefs discussed here. Good luck.
Information technology is also the case where f(a) is non defined but the overall limit equally x-->a exists. The way i can remove this is by defining the role at x=a.If a part has a removable aperture at [tex] x = a [/tex], that means these things.[tex]
\brainstorm{align*}
\lim_{10 \to a^+} f(ten) & = \lim_{ten \to a^-} f(x)\\
\text{ so } & \lim_{x \to a} f(x) \text{ exists}\\
f(a) & \text{ is defined} \\
f(a) & \ne \lim_{x \to a^+} f(x)
\stop{align*}
[/tex]luck.
P.S. I am sure you know this, simply this is adressed to the OP.
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