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China Mathematical Competition

Algebra Difficulty 3.8 AMC 10/12 Find the answer China

The interval on which the function $f(x) = \log_{\frac{1}{2}}(x^2 - 2x - 3)$ is monotone increasing is:

Pick one

Solution

First, we will find the domain of $f(x)$ . From $x^2 - 2x - 3 > 0$ , we obtain $x < -1$ or $x > 3$ . So the domain of definition for $f(x)$ is $(-\infty, -1) \cup (3, +\infty)$ .

But $u = x^2 - 2x - 3 = (x-1)^2 - 4$ is monotone decreasing on $(-\infty, -1)$ , and monotone increasing on $(3, +\infty)$ .

So $f(x) = \log_{\frac{1}{2}}(x^2 - 2x - 3)$ is monotone increasing on $(-\infty, -1)$ , and monotone decreasing on $(3, +\infty)$ .

Answer: A.

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