Question

Question #86e50

Answer 1

`f(x) = | sqrt (x + 9) |`

Explanation:

`f[g(x)] = | 2 x - 3 |`, `g(x) = 4 x^2 -12 x`

`f[4 x^2 -12 x] = | 2 x - 3 |` `->a`

Assuming
`y = 4 x^2 -12 x` `->f[4 x^2 -12 x] = f(y)`
`y = 4 (x^2 -3 x)`
`y/4 = (x^2 -3 x)`
solve using comleting a square
`y/4 = (x -3/2)^2 -(3/2)^2`

`y/4 + 9/4= (x -3/2)^2`

`+- sqrt (y/4 + 9/4)= x -3/2`

`3/2 +-sqrt (y + 9)/2 = x`

`(3 +-sqrt (y + 9))/2 = x`

plug `y` and `x = (3 +-sqrt (y + 9))/2` in se `a`

`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) |`