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Functions:

In image 1: Evaluate the functions

In image 2: Given the functions, find their inverse, if possible and if not possible, redefine Domain or Range

Functions In image 1 Evaluate the functions In image 2 Given the functions find their inverse if possible and if not possible redefine Domain or Range class=
Functions In image 1 Evaluate the functions In image 2 Given the functions find their inverse if possible and if not possible redefine Domain or Range class=

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Answer:

Image 1

1. x = 4

  • f(4) = (- 3*4 + 1) / 6 = -11/6

2. x = -2/3

  • f(-2/3) = -3*(-2/3) + 10 = 2 + 10 = 12

3. x = -5

  • f(-5) = [tex]\sqrt{-(-5)/20}[/tex] = [tex]\sqrt{5/20}[/tex] = [tex]\sqrt{1/4}[/tex] = 1/2

Image 2

  • To work out the inverse, substitute f(x) with x and x with y, then solve for y. The new function is the inverse of the given.

1. f(x) = 5x + 3

  • x = 5y + 3 ⇒ 5y = x - 3 ⇒ y = (x - 3)/5
  • f⁻¹(x) =  (x - 3)/5

2. f(x) = 4/5x

  • x = 4/5y ⇒ y = 5/4x
  • f⁻¹(x) = 5/4x

3. f(x) = (2x + 7)/4

  • x = (2y + 7)/4 ⇒ 2y + 7 = 4x ⇒ 2y = 4x - 7 ⇒ y = 2x - 3.5
  • f⁻¹(x) = 2x - 3.5

4. f(x) = (2x + 7) /(x + 3) = 2 + 1/(x + 3)

  • x = 2 +  1/(y + 3) ⇒ x - 2 = 1/(y + 3) ⇒ y + 3 = 1/(x - 2) ⇒ y = (1 - 3x + 6)/(x - 2) ⇒ y = (7 - 3x)/(x - 2)
  • f⁻¹(x) =  (7 - 3x)/(x - 2)

5. f(x) = [tex]\sqrt{x - 4}[/tex]

  • x = [tex]\sqrt{y - 4}[/tex] ⇒ x² = y - 4 ⇒ y = x² + 4
  • f⁻¹(x) = x² + 4

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