How many discontinuities does the following piecewise function have?

math expressionf left parenthesis x right parenthesis equals enlarged left brace start layout 1st row 3 e superscript x plus 3 baseline plus 1 comma x less than negative 3 2nd row start fraction 2 over 3 end fraction x squared minus 1 comma negative 3 less than x less than or equals 3 3rd row negative start fraction 7 over 2 end fraction x plus start fraction 29 over 2 end fraction comma 3 less than x less than 5 4th row log left parenthesis 2 x minus 4 right parenthesis comma x greater than or equals 5 end layout
(1 point)
Responses

three
three

two
two

one
one

zero

1 answer

To determine the number of discontinuities in the given piecewise function, we need to check the points at which the function changes its definition, specifically at x=3, x=3, and x=5. We will assess the continuity of the function at those points.

  1. At x=3:

    • From the left (x<3): f(x)=3ex+1
      • f(3)=3e3+1 (a specific value).
    • From the right (3<x3): f(x)=23x21
      • f(3)=23(3)21=2391=61=5.
    • Since 3e3+15, there is a discontinuity at x=3.
  2. At x=3:

    • From the left (3<x3): f(x)=23x21
      • f(3)=23(3)21=2391=61=5.
    • From the right (3<x<5): f(x)=72x+292
      • f(3)=72(3)+292=212+292=82=4.
    • Since 54, there is a discontinuity at x=3.
  3. At x=5:

    • From the left (3<x<5): f(x)=72x+292
      • f(5)=72(5)+292=352+292=62=3.
    • From the right (x5): f(x)=log(2x4)
      • f(5)=log(254)=log(104)=log(6).
    • Since 3log(6), there is a discontinuity at x=5.

Summary:

  • The function has discontinuities at x=3, x=3, and x=5.

Thus, the total number of discontinuities is 3. The correct response is:

three

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