Tentukan fungsi invers dari fungsi berikut = A.[tex]f(x) = \sqrt{16-x ^{2} } [/tex] B.[tex]f(x) = 2^{x} [/tex] C.[tex]f(x) = \sqrt{x} +1[/tex]

jawab

a. f(x) = √(16 - x²)
f⁻¹ { √(16 - x²) } = x
y = √(16 -x²)
y² = 16 - x²
x² = 16 - y²
x  =  √(16 - y²)
f⁻¹ (y) = √(16 - y²)
f⁻¹ (x) = √(16 - x²)

b.  f(x) = 2ˣ
f⁻¹ (2ˣ) = x
2ˣ = y
xlog 2  = log y
x = ²log y
f⁻¹(y) = ²log y
f⁻¹ (x) = ²log x

c. f(x) = √x + 1
f⁻¹ (√x + 1) = x
√x + 1 = y
√x = y - 1
x  = (y-1)²
f⁻¹ (y) = (y-1)²
f⁻¹(x ) = (x - 1)²

Kelas 10 Matematika
Bab Fungsi

a] f(x) = √(16 - x²)
y = √(16 - x²)
y² = √(16 - x²)²
y² = 16 - x²
x² = 16 - y²
x = √(16 - x²)
f^(-1) (x) = √(16 - x²)

b] f(x) = 2^x
y = 2^x
x = ²log y
f^(-1) (x) = ²log x

c] f(x) = (√x) + 1
y = (√x) + 1
√x = y - 1
(√x)² = (y - 1)²
x = y² - 2y + 1
f^(-1) (x) = x² - 2x + 1