Question #86e50

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Answer 1 · 2017-03-28T09:48:08

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

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

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

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,
$f(y) = | sqrt (y + 9) |$
$f(x) = | sqrt (x + 9) |$