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Finding value of c to make left/right continuous

  • Thread starter phrox
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  • #1
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Homework Statement


Given:
g(x) = {x+3 for x<-1, cx for -1≤x≤2, x+2 for x>2

Questions:
Find value of c such that g(x) is a)left continuous. b)right continuous.


Homework Equations


??


The Attempt at a Solution


I tried using a method that used finding a point such as (-1,x) and (2,x) but seeing how c is in the middle equation, that didn't work out so well. So I'm stuck as to what to do.
I've done a question sort of like this a long time ago but it was just asking for f(x) to be continuous. Didn't specifically ask about left/right so I'm stumped there.
 

Answers and Replies

  • #2
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There is one value of x where "left continuous" can be wrong, and one point where "right continuous" can be wrong. Can you find those two points?
For (a) you have to care about the first one, for (b) the second one is relevant.
 
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  • #3
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Actually no I don't know how to find those two points sadly.. And what do you mean about you have to care about the first one, and for b the second one is relevant
 
  • #4
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Is the function left-continuous at x=-4?
Is the function left-continuous at x=0?

Is the function left-continuous within those 3 regions of the function definition?
Is the function left-continuous at the first border between those regions?
Is the function left-continuous at the second border between those regions?
Is there any x-value not covered with these three questions?

If one of those answers depend on c, which value of c do you need?
 
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  • #5
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Since x+3 is valid at x=-4, isn't it left continuous?
How do I check for continuity at x=0? Since it says the function is at cx

I guess the main question for me is how am I supposed to find out how to check if its continuous from a side.
 
  • #6
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Since x+3 is valid at x=-4, isn't it left continuous?
"valid"?
At that point, it is continuous, which implies left continuous.

How do I check for continuity at x=0? Since it says the function is at cx
With the definition of continuity.

I guess the main question for me is how am I supposed to find out how to check if its continuous from a side.
As with from both sides, you can just ignore function values on one side (so it is easier in some cases).
 
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  • #7
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Okay, so if the function is continuous until x<-1,to check for this left continuity do I plug -1 into the first equation, giving 2. And then do I plug in -1 into the second equation giving -c? But 2 =/= -c.

That's all I'm getting from this, sorry I don't understand it very well..
 
  • #8
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But 2 =/= -c.
Why? What prevents -c to be 2?
 
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  • #9
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Well I just thought that if 2 is the answer to the first, and -c is the answer to the second, then since they aren't the exact same values, how can they be continuous
 
  • #10
CAF123
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Okay, so if the function is continuous until x<-1,to check for this left continuity do I plug -1 into the first equation, giving 2. And then do I plug in -1 into the second equation giving -c? But 2 =/= -c.

That's all I'm getting from this, sorry I don't understand it very well..
Sometimes it helps to actually sketch the function. Sketch y=x+3 for x<-1 and y=x+2 for x>2. You can leave the interval -1<x<2 clear for just now, but if you drew the sketch accurately, it should quickly help you determine the sign of c.
 
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  • #11
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yes I see the sign of c is positive. How can I just find the value of c from this? Is it just 1?
 
  • #12
CAF123
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yes I see the sign of c is positive.
Yes Edit: misread question, sign of c is clear depending on whether you consider the left continuity or right continuity.
How can I just find the value of c from this? Is it just 1?
It is not what I get. Since each function in each interval are linear functions or polynomials of degree 1, they are continuous on the real line. In particular, ##\lim_{x \rightarrow a} f(x) = f(a)## Use your sketch to decide what 'a' to use and what function to use in each interval.
 
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  • #13
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So I used a=-1 for the x+3 interval, which makes 2. I just don't understand where I'm supposed to go from this step. That "simple" limit you showed is odd to me, the f(a) doesn't make sense to me I guess.
 
  • #14
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If the limit of the first interval is 2, wouldn't this mean that the c value would have to be 2?
 
  • #15
CAF123
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So I used a=-1 for the x+3 interval, which makes 2.
Apologies, I completely misread the question at first. Ignore my previous comment about the sign of c. What you have found above is that ##\lim_{x \rightarrow -1^-} x+3 = 2##. What you need to find in a) is the value of c that makes g(x) right continuous. This means as you approach -1 from the right, the value of cx (we use cx here since g(x) = cx for -1<x<2) tends to 2.
 
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  • #16
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Alright so I clarified that 2 is the answer for a). That part makes sense now. But for b), how would the answer be any different? Since the lines will be in the same spots either way, would it be -2?
 
  • #17
CAF123
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Alright so I clarified that 2 is the answer for a). That part makes sense now. But for b), how would the answer be any different? Since the lines will be in the same spots either way, would it be -2?
Yes, thats correct, the lines both go through the origin if that is what you mean. I am not sure how you made your deduction though. For b), you want to find the value of c that makes ##\lim_{x \rightarrow -1^+} cx = \lim_{x \rightarrow -1^-} x + 3= 2##
 
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  • #18
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What do you mean by thats correct, which part? The answer being any different or the -2? So b) is either 2 or -2?
 
  • #19
CAF123
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What do you mean by thats correct, which part? The answer being any different or the -2? So b) is either 2 or -2?
I mean that both answers are correct, that is c is -2 for (b) and 2 for (a), but I didn't quite understand your reasoning for (b).
 
  • #20
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Oh well I just assumed that since it's left continuous at 2 going in a positive direction, then if it's going in a negative direction(to the right) then it would have to be -2.
 
  • #21
CAF123
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Oh well I just assumed that since it's left continuous at 2 going in a positive direction, then if it's going in a negative direction(to the right) then it would have to be -2.
No, because if say g(x) = 0 for x <-1, then c would obviously be zero to make g right continuous.
 
  • #22
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Oh.. Not sure how it's done then
 
  • #23
CAF123
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Oh.. Not sure how it's done then
Do you understand the notation ##\lim_{x\rightarrow a} f(x) = f(a)##? This is the condition that has to be satisfied for f to be continuous. All functions within the piece wise defined function g(x) are polynomials and and the limit as x tends to some value a is the value of the polynomial at a.

If the equations above holds, then the left and right limits are the same. Now refer to your sketch. The limit of x+3 as x tends to -1 from the left is 2. To make g right continuous, the limit as x tends to -1 from the right should also be 2. But as g tends to -1 from the right, it is described by cx. So you want c such that ##\lim_{x \rightarrow -1^+} cx = 2.## Now solve for c.

Apply the same idea for the point a=2. Does this help?
 

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