The problem provides two points on a linear function $f(x)$: $(3, 2)$ and $(5, 3)$. The goal is to determine the equation for $f(x)$ from the given options. The options are: (a) $f(x) = \frac{1}{2}x - \frac{1}{2}$ (b) $f(x) = \frac{1}{2}x + \frac{1}{2}$ (c) $f(x) = 2x - 1$ (d) $f(x) = 2x + 1$

1. Problem Description

The problem provides two points on a linear function $f(x)$: $(3, 2)$ and $(5, 3)$. The goal is to determine the equation for $f(x)$ from the given options. The options are:

(a) $f(x) = \frac{1}{2}x - \frac{1}{2}$

(b) $f(x) = \frac{1}{2}x + \frac{1}{2}$

(c) $f(x) = 2x - 1$

(d) $f(x) = 2x + 1$

2. Solution Steps

Since we have two points, we can find the slope $m$ of the line:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 2}{5 - 3} = \frac{1}{2}$

Now we have the slope, we can write the equation of the line in point-slope form using the point $(3, 2)$:

$y - y_1 = m(x - x_1)$

$y - 2 = \frac{1}{2}(x - 3)$

$y - 2 = \frac{1}{2}x - \frac{3}{2}$

$y = \frac{1}{2}x - \frac{3}{2} + 2$

$y = \frac{1}{2}x - \frac{3}{2} + \frac{4}{2}$

$y = \frac{1}{2}x + \frac{1}{2}$

So, $f(x) = \frac{1}{2}x + \frac{1}{2}$.

Now let's check if this equation holds true for both points.

For $(3, 2)$:

$f(3) = \frac{1}{2}(3) + \frac{1}{2} = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$

For $(5, 3)$:

$f(5) = \frac{1}{2}(5) + \frac{1}{2} = \frac{5}{2} + \frac{1}{2} = \frac{6}{2} = 3$

The equation $f(x) = \frac{1}{2}x + \frac{1}{2}$ holds true for both points.

3. Final Answer

(b) $f(x) = \frac{1}{2}x + \frac{1}{2}$