If $h (x) = \frac{f (x)}{g (x)}$ $h(x)=(f(x))/(g(x))$ and $f (x) = 3 x - 1$ $f(x)=3x-1$ , how do you determine g(x) given $h (x) = \frac{3 x - 1}{\sqrt{x + 6}}$ $h(x)=(3x-1)/sqrt(x+6)$ ?

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Answer 1 · 2017-02-10T00:57:07

$g (x) = \sqrt{x + 6}$ $g(x) = sqrt(x + 6)$

Set up an equation:

$\frac{3 x - 1}{\sqrt{x + 6}} = \frac{3 x - 1}{g (x)}$ $(3x- 1)/sqrt(x+ 6) = (3x - 1)/g(x)$

$g (x) = \frac{(3 x - 1) (\sqrt{x + 6})}{3 x - 1}$ $g(x) = ((3x- 1)(sqrt(x + 6)))/(3x- 1)$

$g (x) = \sqrt{x + 6}$ $g(x) = sqrt(x+ 6)$

Hopefully this helps!

If h(x)=f(x)g(x)h(x)=(f(x))/(g(x)) and f(x)=3x−1f(x)=3x-1, how do you determine g(x) given h(x)=3x−1√x+6h(x)=(3x-1)/sqrt(x+6)?