Given that $f (x) = \sqrt{x} - 3$ $f(x) = sqrtx - 3$ and g(x) = 2x + 1, how do you find $(\frac{f}{g}) (- \sqrt{3})$ $(f/g)(-sqrt3)$ ?

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Given that $f (x) = \sqrt{x} - 3$ $f(x) = sqrtx - 3$ and g(x) = 2x + 1, how do you find $(\frac{f}{g}) (- \sqrt{3})$ $(f/g)(-sqrt3)$ ?

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Answer 1 · 2017-10-02T13:19:35

See a solution process below:

Explanation:

First, we can write $(\frac{f}{g}) (x)$ $(f/g)(x)$ as:

$(\frac{f}{g}) (x) = \frac{\sqrt{x} - 3}{2 x + 1}$ $(f/g)(x) = (sqrt(x) - 3)/(2x + 1)$

To find $(\frac{f}{g}) (- \sqrt{3})$ $(f/g)(-sqrt(3))$ we need to substitute $- 3$ $color(red)(-3)$ for each occurrence of $x$ $color(red)(x)$ in $(\frac{f}{g}) (x)$ $(f/g)(x)$ :

$(\frac{f}{g}) (x) = \frac{\sqrt{x} - 3}{2 x + 1}$ $(f/g)(color(red)(x)) = (sqrt(color(red)(x)) - 3)/(2color(red)(x) + 1)$ becomes:

$(\frac{f}{g}) (- \sqrt{3}) = \frac{\sqrt{- \sqrt{3}} - 3}{(2 \cdot - \sqrt{3}) + 1}$ $(f/g)(color(red)(-sqrt(3))) = (sqrt(color(red)(-sqrt(3))) - 3)/((2 * color(red)(-sqrt(3))) + 1)$

$(\frac{f}{g}) (- \sqrt{3}) = \frac{\sqrt{- \sqrt{3}} - 3}{(- 2 \sqrt{3} + 1)}$ $(f/g)(color(red)(-sqrt(3))) = (sqrt(color(red)(-sqrt(3))) - 3)/((-2color(red)(sqrt(3)) + 1)$

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Given that $f (x) = \sqrt{x} - 3$ $f(x) = sqrtx - 3$ and g(x) = 2x + 1, how do you find $(\frac{f}{g}) (- \sqrt{3})$ $(f/g)(-sqrt3)$ ?

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Given that f(x)=√x−3f(x) = sqrtx - 3 and g(x) = 2x + 1, how do you find (fg)(−√3)(f/g)(-sqrt3)?

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Given that $f (x) = \sqrt{x} - 3$ $f(x) = sqrtx - 3$ and g(x) = 2x + 1, how do you find $(\frac{f}{g}) (- \sqrt{3})$ $(f/g)(-sqrt3)$ ?