How do you find the domain for $f (x) = \frac{1}{\sqrt{3 - 2 x}}$ $f(x) = 1/(sqrt(3-2x))$ ?

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How do you find the domain for $f (x) = \frac{1}{\sqrt{3 - 2 x}}$ $f(x) = 1/(sqrt(3-2x))$ ?

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Answer 1 · 2016-06-04T02:42:14

Whenever you are dealing with domain/range of a function that involves a radical, you need to set the radical to the following: $x - b \geq 0$ $x - b >= 0$ , since a radical is undefined in the real number system if $x < 0$ $x < 0$ . Once you have set it in the way explained above you solve the resulting inequality.
$\sqrt{3 - 2 x} \geq 0$ $sqrt(3 - 2x) >= 0$

${(\sqrt{3 - 2 x})}^{2} \geq 0^{2}$ $(sqrt(3 - 2x))^2 >= 0^2$

$3 - 2 x \geq 0$ $3 - 2x >= 0$

$3 \geq 2 x$ $3 >= 2x$

$\frac{3}{2} \geq x$ $3/2 >= x$

Interestingly, this is also a rational function. Therefore, we must determine any vertical asymptotes. Vertical asymptotes in a rational function can be found by setting the denominator to 0 and solving for x. Doing this:

$\sqrt{3 - 2 x} = 0$ $sqrt(3 - 2x) = 0$

${(\sqrt{3 - 2 x})}^{2} = 0^{2}$ $(sqrt(3 - 2x))^2 = 0^2$

$3 - 2 x = 0$ $3 - 2x = 0$

$x = \frac{3}{2}$ $x = 3/2$

Therefore, $x \neq \frac{3}{2}$ $x != 3/2$ in this function.

In the domain statement, we can say that $x < \frac{3}{2}$ $x < 3/2$ , because we had no choice but to remove the $x = \frac{3}{2}$ $x = 3/2$ , since this was the equation of the vertical asymptote.

Hopefully this makes sense!

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How do you find the domain for $f (x) = \frac{1}{\sqrt{3 - 2 x}}$ $f(x) = 1/(sqrt(3-2x))$ ?

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