Question

Tolong bantu jawab nomer 4 dan 5

Answer 1

4.  f(x) = x² - 4x + 1
x² - 4x +1 = y
x² - 4x = y-1
(x-2)² = y -1 + 4  = y + 3
x - 2 = +- √(y+3)
x = 2 +- √(y+3)
f⁻¹(x) = 2 +- √(x+3)

f⁻¹(3-x) = 2 +- √(3-x+3)
f⁻¹(3-x) = 2 +- √(6-x)

5) f(x) = x² - 2x + 5
f⁻¹ (x² -2x + 5) = x

a) f⁻¹(5) ! = 
x²-2x+5 = 5
x²-2x = 0
x(x-2)= 0
x=0 atau x = 2
f⁻¹(5)! = 0 ! = 1
f⁻¹(5)! = 2! = 2

b) f⁻¹(13) ! = 
x² - 2x + 5 = 13
x² -2x - 8 = 0
(x-4)(x+2) = 0
x = 4 atau x = -2 (TM)
f⁻¹(13)! = 4 ! = 4 x 3 x 2 x 1= 24

c) f⁻¹(x) = 2
f⁻¹(x²-2x+5) = x
x² -2x + 5 = y
x² - 2x = y - 5
(x -1)² = y-5+1 = y - 4
(x -1) = √(y-4)
x = 1 + √(y-4)
f⁻¹(x) = 1 + √(x-4)
f⁻¹(x) = 2
1 + √(x-4) = 2
+_ √(x-4) = 1
+- (x- 4) = 1
x - 4 = 1  atau 4 - x = 1
x = - 5 atau x = 3

d) f⁻¹(x) = 1
1 +_ √(x-4) = 1
1 + √(x-4) = 1  atau 1 - √(x-4) = 1
√(x-4) = 0  atau √(x-4) = 0
x = 4