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Sergeu [11.5K]
3 years ago
14

Which values of x are point(s) of discontinuity for this function? Please explain how you did it.

Mathematics
2 answers:
solniwko [45]3 years ago
5 0

Answer:SAME AS ABOVE

Step-by-step explanation:

bixtya [17]3 years ago
3 0
<span>The correct answers are:
x = 0
x = 2

Explanation:
When x < 0, the value of y = -</span>x^2<span>.
When x > 0, the value of y = </span>x^2<span>.
When x = 0, there is no value of y mentioned in the conditions; it means that there is discontinuity at x = 0.

When x < 2, the value of y = </span>x^2<span>.
When x >= 2, the value of y = 2.

It means that, at x = 2, there is a discontinuity because the graph of y = </span>x^2 is not touching the graph of y = 2 at x = 2.
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I need some help with this calculus 1 implicit diffrention problem?
katen-ka-za [31]

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