Reach for the Summit - M-S1-A3

Note that the function $y = \log_\frac 12 (x^2+4x+4) = \log_\frac 12 (x+2)^2$ has a domain of $x \in (-\infty, -2) \cup (-2, \infty)$ .

Let us check the gradient of $y$ .

$\begin{aligned} y & = \log_\frac 12 (x+2)^2 = \frac {2\ln (x+2)}{- \ln 2} \\ \implies \frac {dy}{dx} & = \frac {-2}{\ln 2(x+2)} \end{aligned}$

$\dfrac {dy}{dx} > 0$ and hence function $y$ is increasing when $\boxed{(-\infty, -2)}$ .

$y=f(x)=\log_{0.5} (x^2+4x+4)$

$=-\dfrac {2\ln (x+2)}{\ln 2}$

$\implies f'(x) =-\dfrac {2}{(x+2)\ln 2}>0$

$\implies x<-2$

So the required interval is $(-\infty, -2$ ).