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Domain of $f(x) < 3$ is $(0,\infty)$ and

Domain of $f(x) > -1$ is $(-\infty , 5)$

What is the domain of $[f(x)]^2 \ge f(x) + 6$?

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This question does not show any effort. –  user2468 May 14 '12 at 19:50
HINT: $[f(x)]^2-f(x)-6=(f(x)-3)(f(x)+2)$. –  Brian M. Scott May 14 '12 at 19:50
@BrianM.Scott Can you please tell how to further solve the question? –  Sk D Champ May 14 '12 at 19:53
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closed as too localized by Jennifer Dylan, Asaf Karagila, tomasz, William, rschwieb Aug 31 '12 at 12:34

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1 Answer

$[f(x)]^2\ge f(x)+6$ if and only if $[f(x)]^2-f(x)-6\ge 0$. Now use the fact that $$[f(x)]^2-f(x)-6=\Big(f(x)-3\Big)\Big(f(x)+2\Big)\;,$$ remembering that for any real numbers $a$ and $b$, $ab=0$ if and only if at least one of $a$ and $b$ is $0$, and $ab>0$ if and only if either $a>0$ and $b>0$, or $a<0$ and $b<0$. Use the information that you were given to decide where $f(x)-3$ and $f(x)+2$ are are positive or negative.

However, I think that there is a typographical error in the statement of the problem: the second condition should be that $x>-2$ on $(-\infty,5)$ in order for you to work the problem.

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