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Find the numbers at which $ f $ is discontinuous. At which of these numbers is $ f $ continuous from the right, from the left, or neither? Sketch the graph of $ f $.

$ f(x) = \left\{ \begin{array}{ll} x^2 & \mbox{if $ x < -1 $}\\ x & \mbox{if $ -1 \le x < 1 $} \\ 1/x & \mbox{if $ x \ge 1 $} \end{array} \right.$

\[\text { Now } \lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}} x^{2}=1 \text { and } \lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{-}} x=-1\]so $f$ is discontinuous at $-1 .$ since $f(-1)=-1, f$ is continuous from the right at -1. Also, $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} x=1$ and $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} \frac{1}{x}=1=f(1),$ so $f$ is continuous at 1.

04:53

Daniel J.

01:34

Tyler M.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

Missouri State University

Baylor University

University of Michigan - Ann Arbor

Boston College

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Okay. This question is asking about continuity of this function F. And so the only possible point we know that one over X is continuous when x is bigger than one, X is continuous on the interval negative 1 to 1. And X squared is continuous when x is less than negative one. So the only interesting points are when X equals negative one and when X equals one at X equals negative 12 of the functions are meeting X squared and X there potentially meeting. So I need to look at the limit of X squared on its domain. So as X approaches negative one from the left negative one squared comes out to one and I am approaching x equals negative one but not including. So I'm looking at if this is my vertical axis and here's my horizontal This grid is kind of weird. I'm going to let three spaces B1. So as X approaches negative one from the left, I'm following the graph of Y equals X squared. So I'm approaching -11 and negative two. I'm up at four. So 1234 be way up here somewhere and the graph will look like this. But it has an open circle at this point. As X approaches negative one from the left, X squared approaches positive one. But since X is not equal to negative one in the domain of that function, I can't finish the doctor. Now I'm looking at Michael's ex limit as X. Sorry, Y equals X Limit as X approaches -1 from the right As X approaches -1 from the right of X. I'll get -1. This stopped, so I'm following the line Y equals X. Starting at -1 equal to negative one and up to positive one. But not including Now back to here. Since the limit as X approaches -1 from the left And the limited x approaches negative one from the right are not equal. Ex's D F of X is discontinuous at X equals negative one. They're approaching two different values. Now I'm right side continuous as X approaches negative one of Y equals X because my limit as X approaches negative one from the right is equal to function value. But I don't have a continuity at negative one between the two branches of the graph. I'm gonna look at x equals positive one. I wrote it up here but we're going to move it down to hear And look at x equals one. The limit limit of X as X approaches one from the left is one but I don't go all the way up in including I go up to an open circle. The limit of one over X as X approaches one from the right is also one. So f is continuous at X equals one because the limit from the left is equal the limit from the right one over X is gonna look like one and then 1/2 than one third one over X. When x is bigger than one looks like the blue graph. So I'm continuous at one and discontinuous at -1, Discontinues -1 because the left hand is not equal the right hand there, and I'm continuous at one because the left hand equals the right hand. Thanks for listening.

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