kaysimon17
kaysimon17 kaysimon17
  • 10-11-2018
  • Mathematics
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Evaluating a piece wise defined function

Evaluating a piece wise defined function class=

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Rinkhals
Rinkhals Rinkhals
  • 10-11-2018

Given : if x belongs to the interval -2 ≤ x < 2 then h(x) = -(x - 1)² + 1

As x = -2 belongs to the interval -2 ≤ x < 2

⇒ h(-2) = -(-2 - 1)² + 1 = -(-3)² + 1 = -9 + 1 = -8

As x = 1 belongs to the interval -2 ≤ x < 2

⇒ h(1) = -(1 - 1)² + 1 = 0 + 1 = 1

Given if x is greater than or equal to 2 then h(x) = [tex]\frac{-1}{4}x + 1[/tex]

As x = 3 is greater than 2

⇒ h(3) = [tex]\frac{-1}{4}(3) + 1 = (\frac{-3}{4} + 1) = \frac{1}{4}[/tex]

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