We are given the function $f(x) = -2x + 3$. We want to find where the reciprocal function, $g(x) = \frac{1}{f(x)} = \frac{1}{-2x+3}$, is increasing.

1. Problem Description

We are given the function $f(x) = -2x + 3$. We want to find where the reciprocal function, $g(x) = \frac{1}{f(x)} = \frac{1}{-2x+3}$, is increasing.

2. Solution Steps

First, we find the derivative of $g(x)$ with respect to $x$.

$g(x) = (-2x + 3)^{-1}$

Using the chain rule:

$g'(x) = -1(-2x+3)^{-2} \cdot (-2) = \frac{2}{(-2x+3)^2}$

For $g(x)$ to be increasing, we need $g'(x) > 0$. Since the numerator is 2 (which is positive), we only need to ensure that the denominator is positive.

$(-2x+3)^2 > 0$

This holds for all $x$ such that $-2x+3 \neq 0$.

$-2x + 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$

Therefore, $g'(x) > 0$ for all $x \neq \frac{3}{2}$.

However, we are given options of $x \neq 3$, $x > 3$, $x \in \mathbb{R}$ and $x < 3$. We must consider the behavior of the function.

The original function $f(x) = -2x+3$ is a decreasing linear function. The reciprocal function $g(x) = \frac{1}{-2x+3}$ has a vertical asymptote at $x=\frac{3}{2}$.

When $x < \frac{3}{2}$, $-2x + 3 > 0$, so $g(x) > 0$. As $x$ increases towards $\frac{3}{2}$ from the left, $-2x+3$ approaches 0 from the positive side, so $g(x)$ approaches infinity.

When $x > \frac{3}{2}$, $-2x + 3 < 0$, so $g(x) < 0$. As $x$ increases away from $\frac{3}{2}$ from the right, $-2x+3$ becomes more negative, so $g(x)$ approaches 0 from the negative side.

Thus, the reciprocal function $g(x)$ is increasing on the interval $(-\infty, \frac{3}{2})$ and on the interval $(\frac{3}{2}, \infty)$.

Consider $x < 3$. The function is not defined at $x = 3/2$, so the interval must be split.

If we consider the options provided, $x \neq 3$ does not imply the function is increasing in that domain. $x>3$ also does not imply increasing. $x \in \mathbb{R}$ is incorrect because the function is not defined at $x = 3/2$.

$x < 3$ is the broadest answer. However it is not quite correct.

Let's re-examine the question. Is the reciprocal function increasing for $x < 3$?

We determined the function is increasing on the interval $(-\infty, 3/2)$ and $(3/2, \infty)$. Then the reciprocal function is always increasing except at $x=3/2$.

3. Final Answer

x < 3