Question #86e50

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Question #86e50

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Answer 1 · 2017-03-28T09:48:08

$f (x) = ∣ ∣ \sqrt{x + 9} ∣ ∣$ $f(x) = | sqrt (x + 9) |$

Explanation:

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

$f [4 x^{2} - 12 x] = | 2 x - 3 |$ $f[4 x^2 -12 x] = | 2 x - 3 |$ $\to a$ $->a$

Assuming
$y = 4 x^{2} - 12 x$ $y = 4 x^2 -12 x$ $\to f [4 x^{2} - 12 x] = f (y)$ $->f[4 x^2 -12 x] = f(y)$
$y = 4 (x^{2} - 3 x)$ $y = 4 (x^2 -3 x)$
$\frac{y}{4} = (x^{2} - 3 x)$ $y/4 = (x^2 -3 x)$
solve using comleting a square
$\frac{y}{4} = {(x - \frac{3}{2})}^{2} - {(\frac{3}{2})}^{2}$ $y/4 = (x -3/2)^2 -(3/2)^2$

$\frac{y}{4} + \frac{9}{4} = {(x - \frac{3}{2})}^{2}$ $y/4 + 9/4= (x -3/2)^2$

$\pm \sqrt{\frac{y}{4} + \frac{9}{4}} = x - \frac{3}{2}$ $+- sqrt (y/4 + 9/4)= x -3/2$

$\frac{3}{2} \pm \frac{\sqrt{y + 9}}{2} = x$ $3/2 +-sqrt (y + 9)/2 = x$

$\frac{3 \pm \sqrt{y + 9}}{2} = x$ $(3 +-sqrt (y + 9))/2 = x$

plug $y$ $y$ and $x = \frac{3 \pm \sqrt{y + 9}}{2}$ $x = (3 +-sqrt (y + 9))/2$ in se $a$ $a$

$f(y) = | cancel2 ((3 +-sqrt (y + 9))/cancel2) - 3|$

$f(y) = | 3 +-sqrt (y + 9) - 3|$

$f(y) = | +-sqrt (y + 9) |$

since it result always +ve value (absoluted). Therefore,
$f(y) = | sqrt (y + 9) |$
$f(x) = | sqrt (x + 9) |$

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Question #86e50

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