Find range of the following function.

Question

Find the range of

$$f(x)=\log_2(\log_{\frac12} (x^2+4x+4))$$

Answer 1

$$f(x)=Log(-2 \log((x+2)^2/2) \Rightarrow -2 \log((x+2)^2/2) >0 \Rightarrow \log ((x+2)^2/2) < \log 1. $$ So for the domain of the function, we have $$(x+2)^2 <1/2 \Rightarrow -2-\frac{1}{\sqrt{2}} < x < -2+\frac{1}{\sqrt{2}}$$ Finally ,the domain of $F(x)$ is $(-2-\frac{1}{\sqrt{2}}, -2) \cup (-2, -2+\frac{1}{\sqrt{2}})$. The right limit of $f(x)$ at $x=-2-\frac{1}{\sqrt{2}}$ and left limit of $f(x)$ at $x=-2+\frac{1}{\sqrt{2}}$ are $-\infty$. at $x=-2$ is $+\infty$ are $-\infty$. Similarly, the left and the right limit of $f(x)$ at $x=-2$ are $+\infty$. So the range of thios function id $y \in(-infty, \infty).$ $$y=\log(- \log (4(x^2+4x+4)^2) \Rightarrow (4(x^2+4x+4) \Rightarrow 4(x+2)^2=\exp({-\exp(y)}>0,$$ so $y\in(-\infty, \infty)$. This confirms therange of $f(x)$. See the fiogure below