Determine whether the function is continuous at the given x-value. If discontinuous, state the type of discontinuity (jump, inf., removable)?

[TranceDreamscape009]

1. y=(3)/(2x^2)
at x=-1

2. y=(x-2)/(x-4)
at x=4

3. y=(x+2)/(x^2+6x+8)
at x=-2

4. y={x^2+1, if x<-3 -x-3, if x≥-3}
at x=2

5. y={x^2+1, if x<-3 -x-3, if x≥-3}
at x=-3

 

[JeffM]

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What is the meaning of

[MATH]x \le - 3 - x - 3 \implies 2x \le -6 \implies x \le -3[/MATH]
which is true given the premise, but of what import is it?

 

[Dr.Peterson]

1. y=(3)/(2x^2)
at x=-1

2. y=(x-2)/(x-4)
at x=4

3. y=(x+2)/(x^2+6x+8)
at x=-2

4. y={x^2+1, if x<-3 -x-3, if x≥-3}
at x=2

5. y={x^2+1, if x<-3 -x-3, if x≥-3}
at x=-3

I think the last function is meant to be y={x^2+1, if x<-3; -x-3, if x≥-3}; that is,

[MATH]y=\left\{\begin{matrix} x^2+1 & \text{if } x< -3\\ -x-3 & \text{if } x\geq -3 \end{matrix}\right.[/MATH]
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