Question

The domain of the function $\log_{10}(x^2 - 5x + 6)$ is

• A. $(-\infty, \infty)$
• B. $(-\infty, 2) \cup (3, \infty)$
• C. $(2, 3)$
• D. None of these

Hint:

For any strict quadratic inequality of the form $(x - \alpha)(x - \beta) > 0$ where $\alpha < \beta$, the valid solution intervals always lie strictly outside the roots, namely $(-\infty, \alpha) \cup (\beta, \infty)$. If it were $< 0$, the solution would lie inside the roots!

Answer 1

The Correct Option is B

Step 1: State the domain condition for a log.
$\log_{10}(\,\cdot\,)$ is defined only when its argument is strictly positive, so we need $x^2 - 5x + 6 > 0$.
Step 2: Factor the quadratic.
$x^2 - 5x + 6 = (x-2)(x-3)$.
Step 3: Locate the critical points.
The product $(x-2)(x-3)$ is zero at $x = 2$ and $x = 3$, which split the number line into three intervals.
Step 4: Test a point in each interval.
For $x = 0$: $(-)(-) = + > 0$. For $x = 2.5$: $(+)(-) = - < 0$. For $x = 4$: $(+)(+) = + > 0$.
Step 5: Keep where the product is positive.
The expression is positive for $x < 2$ and for $x > 3$.
Step 6: Write the domain in interval notation.
$(-\infty, 2) \cup (3, \infty)$.
\[ \boxed{(-\infty, 2)\cup(3, \infty)} \]