Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

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

share|improve this question
2  
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
add comment

closed as too localized by Jennifer Dylan, Asaf Karagila, tomasz, William, rschwieb Aug 31 '12 at 12:34

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.