f[g(x)] = | 2 x - 3 |f[g(x)]=|2x−3|, g(x) = 4 x^2 -12 xg(x)=4x2−12x
f[4 x^2 -12 x] = | 2 x - 3 |f[4x2−12x]=|2x−3| ->a→a
Assuming
y = 4 x^2 -12 xy=4x2−12x ->f[4 x^2 -12 x] = f(y)→f[4x2−12x]=f(y)
y = 4 (x^2 -3 x)y=4(x2−3x)
y/4 = (x^2 -3 x)y4=(x2−3x)
solve using comleting a square
y/4 = (x -3/2)^2 -(3/2)^2y4=(x−32)2−(32)2
y/4 + 9/4= (x -3/2)^2 y4+94=(x−32)2
+- sqrt (y/4 + 9/4)= x -3/2±√y4+94=x−32
3/2 +-sqrt (y + 9)/2 = x 32±√y+92=x
(3 +-sqrt (y + 9))/2 = x 3±√y+92=x
plug yy and x = (3 +-sqrt (y + 9))/2 x=3±√y+92 in se aa
f(y) = | cancel2 ((3 +-sqrt (y + 9))/cancel2) - 3|
f(y) = | 3 +-sqrt (y + 9) - 3|
f(y) = | +-sqrt (y + 9) |
since it result always +ve value (absoluted). Therefore,
f(y) = | sqrt (y + 9) |
f(x) = | sqrt (x + 9) |