TMUA practice: Exponentials Logs, problem 5

Domain. Need $\;u(x) = x^{2} - 2x - 3 = (x - 3)(x + 1) > 0$, i.e. $\;x < -1\;$ or $\;x > 3$.

Monotonicity. $\log_{1/2}$ is a decreasing function. So $f(x) = \log_{1/2}(u(x))$ is decreasing iff $u(x)$ is increasing.

$u'(x) = 2x - 2$, which is positive iff $x > 1$. So $u$ is increasing on $(1, \infty)$.

Intersect with the domain: $(1, \infty) \cap [\,(-\infty, -1) \cup (3, \infty)\,] = (3, \infty)$.

So $f$ is monotonically decreasing on $(3, \infty)$. This is option D.

Why the other options fail.

• A, $(-\infty, -1)$: in this part of the domain $u$ is decreasing (since $x < -1 < 1$), so $f = \log_{1/2}(u)$ is increasing.
• B, $(-\infty, 1)$: includes points outside the domain (those between $-1$ and $1$).
• C, $(1, \infty)$: includes points between $1$ and $3$ where the function is undefined.

The lesson: for $\log_{a} \circ u$ with $0 < a < 1$, the composite is decreasing where $u$ is increasing. Combine with the domain $u > 0$.