## cwtan Group Title Find the range of $$\Huge \frac {3x^2-2x-1}{x^2+x+2}$$ if x is real. one year ago one year ago

1. Jemurray3 Group Title

Is calculus available as a tool for this?

2. cwtan Group Title

I dun think it need calculus.....

3. Jemurray3 Group Title

No, but it would be quite straightforward if you used it.

4. cwtan Group Title

but this is a 4 mark question and i need the steps to get the mark...... if use calculus i dun think i can get full mark for this question.... (at least 1 mark 1 step)

5. Jemurray3 Group Title

okay, well that sounds like there's a particular way you're supposed to do it. What class is it?

6. cwtan Group Title

Polynomial/

7. Jemurray3 Group Title

I'm not sure what you mean by that. Is it an algebra class or something? And does it ask you to complete a series of 4 steps?

8. cwtan Group Title

somtething like: http://www.stpmwiki.com/index.php/Polynomials_Part2

9. cwtan Group Title
10. Jemurray3 Group Title

whoa whoa whoa. Are you looking for a range or a remainder?

11. cwtan Group Title

i duno which part can be use to find the range :( but i know the syllabus is just in that 2 webpage

12. Jemurray3 Group Title

I say again, just to clarify... you're looking for the range of the function? Which means all of the possible values it could take? Or you're looking for the remainder of the fraction once you've divided?

13. cwtan Group Title

find the range for the fraction possible

14. ganeshie8 Group Title

lets say, $$\frac {3x^2-2x-1}{x^2+x+2} = k$$ => $$3x^2-2x-1 = k(x^2+x+2)$$ $$(3-k)x^2 + (-2-k)x + (-1-2k) = 0$$

15. cwtan Group Title

Thank you veryyyy much!!

16. ganeshie8 Group Title

since, x is real, $$b^2 - 4ac >= 0$$ => $$(2+k)^2 - 4(k-3)(1+2k) >= 0$$ $$(k-4)(k+4/7) >= 0$$

17. ganeshie8 Group Title

np :)

18. Jemurray3 Group Title

I like that! I've never done it that way before. @cwtan I'm somewhat surprised that this is a viable technique for you in this instance given what I read in your syllabus, but I suppose if it works it works.

19. ganeshie8 Group Title

yea this is fallback option when calculus is not allowed.... mukushla taught me this few few days back... :)