Find the values that the function f(x) = (x²-3x+2)/2x

Find the values that the function $\displaystyle f(x) = \frac{x^{2}-3x+2}{2x}$ cannot take for any value of the variable.

Step 1: Turn the question into an equation

A number $y$ belongs to the range exactly when $f(x) = y$ has a solution $x\neq0$. Multiplying by $2x$: \[2xy = x^{2}-3x+2 \;\Longleftrightarrow\; x^{2}-\bigl(3+2y\bigr)x+2 = 0. \tag{1}\]

Step 2: The discriminant

\[ \Delta = \bigl(3+2y\bigr)^{2}-8 = 4y^{2}+12y+9-8 = 4y^{2}+12y+1. \tag{2} \]

Equation (1) has a real root precisely when $\Delta\ge0$.

Step 3: Solve $\Delta\ge0$

The roots of $4y^{2}+12y+1 = 0$ are

\[ y = \frac{-12\pm\sqrt{144-16}}{8} = \frac{-12\pm\sqrt{128}}{8} = \frac{-12\pm8\sqrt2}{8} = -\frac32\pm\sqrt2. \tag{3} \]

The parabola opens upward, so $\Delta\ge0$ outside the roots: $y \le -\frac32-\sqrt2 \qquad\text{or}\qquad y \ge -\frac32+\sqrt2$.

Step 4: The excluded value $x = 0$ causes no trouble

Substituting $x = 0$ into the left side of (1) gives $2\neq0$, so $x = 0$ is never a root and no solution has to be discarded.

Step 5: Verification

\[f(x) = \frac{x}{2}-\frac32+\frac1x \;\Longrightarrow\; f'(x) = \frac12-\frac{1}{x^{2}} = 0 \ \text{ at } x = \pm\sqrt2 \;\;\checkmark\]

\[f\bigl(\sqrt2\bigr) = -\frac32+\sqrt2 = -0.0857864 \ (\text{minimum for } x\gt 0); \qquad f\bigl(-\sqrt2\bigr) = -\frac32-\sqrt2 = -2.9142136 \ (\text{maximum for } x\lt 0) \;\;\checkmark\]

\[f(1) = f(2) = 0,\ f(-1) = -3:\ \text{values on either side of the gap} \;\;\checkmark\]