Tolong cari bentuk inversnya

Fungsi InveRs

f(x) = y = x/√(x² - 8)
√(x² - 8) /x = 1/y
Kuadratkan kedua ruas
(x² - 8)/x² = 1/y²
1 - 8/x² = 1/y²
8/x² = 1 - 1/y²
8/x² = (y² - 1)/y²
x² = 8y² / (y² - 1)
x = ± √((8y² /(y² - 1))
x = ± (2y√2) / √(y² - 1)

f^-1 (x) = ± (2x√2) / √(y² - 1)

jawab

y = x/ √(x²-8) -->kuadratkan
y² = x²/(x²- 8)
y²(x² -8) = x²
x² y² - 8y² = x²
x² y² - x²  = 8y²
x² (y² -1) = 8y²
x² = (8y²)/(y² -1)
x = √ [ (8y²)/(y² -1) ]
f⁻¹(x) = √ [ (8 x² ) / ( x² - 1) ]