Continuity on piecewise function

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SUMMARY

The discussion focuses on the continuity of the piecewise function defined as f(x) = -2x for -1 ≤ x < 1, f(x) = -2/(x-1) for 1 < x < 2, and f(x) = x-2 for x > 2. The function is discontinuous at x = 1 and x = 2, both classified as jump discontinuities. At x = -1, the function is right continuous, while it is continuous on the intervals (1, 2) and (2, ∞). The user seeks clarification on the classification of discontinuities and the conditions for continuity.

PREREQUISITES
  • Understanding of piecewise functions
  • Knowledge of limits and continuity
  • Familiarity with types of discontinuities (removable, jump, infinite)
  • Basic calculus concepts, including limit evaluation
NEXT STEPS
  • Study the classification of discontinuities in depth
  • Learn about the epsilon-delta definition of continuity
  • Explore the concept of limits approaching from both sides
  • Practice solving more complex piecewise functions for continuity
USEFUL FOR

Students studying calculus, particularly those focusing on continuity and limits, as well as educators seeking to clarify concepts related to piecewise functions and discontinuities.

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Homework Statement


[10 Marks] At which points is the following function continuous and at which point is it discontinuous. Explain the types of discontinuity at each point where the function is discontinuous. Then at each point of the discontinuity, if possible, find a value for f(x) that makes it continuous or one sided continuous.

f(x) =
-2x if -1[itex]\leq[/itex]x<1
-2/(x-1) if 1<x<2
x-2 if x> 2

Homework Equations



Test continuity at point:
f(a) is defined
lim f(x) exists
x->a
lim f(x) = f(a)
x->a

Continuity at Endpoints
lim f(x) = f(a) = left continuous
x->a-
lim f(x) = f(a) = right continuous
x->a+

The Attempt at a Solution



Im thinking what I need to do, is:
Check for continuity at the points -1, 1, 2.
Then I would classify any discontinuities as either removable, jump, or infinite discontinuities.
But that last part, I am not sure what its asking?
What I have is this so far:

Discontinuous at x=1 and x=2.
At x=1: Jump discontinuity
At x=2: Jump discontinuity

Im not sure if I am supposed to do this or if its right, but I did it anyway:
At x=-1, the function is right continuous on the interval [-1, 1). The function is also continuous on (1, 2) and (2, infinity)
 
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Basically correct but the discontinuity at x= 1 is NOT a "jump" discontinuity:
[tex]\lim_{x\to 1^+} f(x)= \lim_{x\to 1}\frac{-2}{x- 1}= -\infty[/tex]
 
Oh really? Cool! Thanks! So by stating the intervals of continuity, I satisfied the last part of the problem? Cause I was not sure what it was asking..
 

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