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

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

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Answer 1 · 2014-09-27T01:23:32

The natural domain of $f$ $f$ is $(- \infty, \frac{3}{2})$ $(-infty,3/2)$ .

Let us look at some details.

There are two rules you need to keep in mind when it comes to finding the natural domain of a function. Here are the rules:

Rule 1: Nonzero in the denominator
Rule 2: Nonnegative in the square-root (eventh-root)

By Rule 1,
$\sqrt{3 - 2 x} \neq 0 \Rightarrow 3 - 2 x \neq 0 \Rightarrow - 2 x \neq - 3 \Rightarrow x \neq \frac{3}{2}$ $sqrt{3-2x} ne 0 Rightarrow 3-2x ne 0 Rightarrow -2x ne -3 Rightarrow x ne 3/2$
By Rule 2,
$3 - 2 x \geq 0 \Rightarrow - 2 x \geq - 3 \Rightarrow x \leq \frac{3}{2}$ $3-2x ge0 Rightarrow -2x ge -3 Rightarrow x le 3/2$

By combining the two conditions above, $x < \frac{3}{2}$ $x < 3/2$ .

Hence, the natural domain is $(- \infty, \frac{3}{2})$ $(-infty, 3/2)$ .

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

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