Question

What is the range of $f\left(x\right)=\frac{1}{x^2+6x+6}$

• A. $(0,\infty )$
• B. $(-\infty ,\frac{1}{3}]\cup [0,\infty )$
• C. $(-\infty -\frac{1}{3}]\cup (0,\infty )$
• D. $(-\infty ,\frac{1}{3}]\cup (0,\frac{1}{3}]$

Answer 1

The correct option is C $(-\infty -\frac{1}{3}]\cup (0,\infty )$

$x^2$ + 6x + 6 = $x^2$ + 6x + 9 - 3 = $(x+3{)}^2$ - 3
$(x+3{)}^2-3$ has a range $[-3,\infty )$
We need to find the range of $\frac{1}{(x+3{)}^2-3}$
Let us break the range of $(x+3{)}^2-3$ into two ranges [-3, 0) and (0, $\infty$).
When $(x+3{)}^2-3$ ranges from $(0,\infty ),\frac{1}{(x+3{)}^2-3}$ ranges
from (0, $\infty$)
When $(x+3{)}^2-3$ ranges from [-3, 0), $\frac{1}{(x+3{)}^2-3}$ ranges
from $(-\infty ,-\frac{1}{3}]$
Overall range = $(-\infty ,-\frac{1}{3}]\cup (0,\infty )$