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