Question

F(x) = √(x²+1) dan (f o g)(x) = (1/x-2) (√x²-4x+5)
tentukan (f o g)-¹ (x)

Answer 1

  f(x) = √(x² +1)
f⁻¹(x) = √(x² -1)

(fog)(x) = (1/(x-2)).(√(x² - 4x + 5))
             = √((x²-4x+5)/(x-2)²
             = √(((x²-4x+4)+1)/(x-2)²)
             = √((1/(x-2)²) + 1)

                   f(x) = √(x² +1)
              (fog)(x)= f(g(x))
√((1/(x-2)²) + 1) = √(g(x)² + 1)
     (1/(x-2))²+ 1) = (g(x)² + 1)
               1/(x-2) = g(x)
                     g⁻¹(x) = (1+2x)/x
                               = (1/x) + 2

(fog)⁻¹ = (g⁻¹(f⁻¹)(x))
           = (1/x) + 2
           = (1/(f⁻¹(x)) + 2
           = (1/√(x²-1)) + 2
           

Answer 2

Y = (√(x² - 4x+5))/(x-2)
y² (x -2)² = x² - 4x + 5
y²(x²-4x+4) = x² - 4x+ 5
y².x² - 4y² x + 4y² - x² + 4x = 5
y² x² - x²  - 4y²x + 4x = 5 - 4y²
(y² -1) x² - 4x(y²-1) = (5 - 4y²)
x² - 4x  = (5 - 4y²)/(y²-1)
(x-2)² = (5-4y²)/(y²-1) + 4
(x-2)² = (5-4y²+4(y²-1))/ (y²-1)
(x-2)² = 1/(y²-1)
x - 2 = +_ √(1/(y²-1))
.
x = 2 +- √(1/(y²-1)
(fog)⁻¹(x) = 2 +- √(1/(x²-1))