Question

Buktikan bahwa jika f(x) = ax2 + bx + c, maka f^-1 (x) = (-b +/- akar(4ax + D))/2a

Answer 1

f(x) = y = ax² + bx + c
ax² + bx = y - c ---> dibagi a
x² + b/a x = y/a - c/a ---> ditambah ( 1/2 b/a )^2 = 1/4 b²/a²
x² + b/a x + 1/4 b²/a² = y/a - c/a + 1/4 b²/a²
( x + 1/2 b/a )^2 = ( y - c ) / a + b² / ( 4a² )
= 4a( y - c ) / ( 4a^2 ) + b^2 / ( 4a² )
= 4a( y - c ) + b² / ( 4a² )
= [ 4ay - 4ac + b² ] / 4a² ---> b² - 4ac = D
= [ 4ay + D ] / 4a²

( x + 1/2 b/a )² = [ 4ay + D ] / 4a²
x + 1/2 b/a = ± √[ (4ay + D) / 4a² ] - 1/2 b/a
x = ± √(4ay + D) / 2a - b / 2a
= [ -b ± √(4ay + D) ] / 2a ---> x = f^-1(y)

f^-1(y) = [ -b ± √(4ay + D) ] / 2a ---> ganti y dengan x
f^-1(x) = [ -b ± √(4ax + D) ] / 2a ---> TERBUKTI ... !