Question

Find the range of \(\Huge \frac {3x^2-2x-1}{x^2+x+2}\) if x is real.

Answer 1

Is calculus available as a tool for this?

Answer 2

I dun think it need calculus.....

Answer 3

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

Answer 4

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)

Answer 5

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

Answer 6

Polynomial/

Answer 7

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?

Answer 8

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

Answer 9

and
http://www.stpmwiki.com/index.php/Polynomials_Part1

Answer 10

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

Answer 11

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

Answer 12

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?

Answer 13

find the range for the fraction possible

Answer 14

lets say,

\(\frac {3x^2-2x-1}{x^2+x+2} = k \)

=>

\(3x^2-2x-1 = k(x^2+x+2)\)

\((3-k)x^2 + (-2-k)x + (-1-2k) = 0\)

Answer 15

Thank you veryyyy much!!

Answer 16

since, x is real, \(b^2 - 4ac >= 0 \)

=>

\((2+k)^2 - 4(k-3)(1+2k) >= 0\)

\((k-4)(k+4/7) >= 0\)

Answer 17

np :)

Answer 18

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.

Answer 19

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