## anonymous one year ago Whats the domain and range for f(x)= x^2+x-2/x^2-3x-4

1. arindameducationusc

Well for domain, the roots of the denominator should not be zero.

2. anonymous

Is the domain all real numbers except for -1, and 4?

3. anonymous

and is the range 1?

4. anonymous

so the numerator and denominator are 2 different domains?

5. UsukiDoll

the domain is in the x-axis and the range is in the y-axis. Since we are given a fraction as a function there is going to be restrictions in the domain.

6. UsukiDoll

$(x^2-3x-4) \rightarrow (x+1)(x-4)$ so the domain should be all real numbers except when x = -1 and x = 4. which is what I saw earlier. so yes you're correct on that one @Abbs__

7. UsukiDoll

that doesn't sound right at all ^

8. UsukiDoll

@Michele_Laino

9. anonymous

Thanks!

10. arindameducationusc

Yes Yes, I did a calculation mistake, @Abbs_ is right

11. anonymous

It's okay. :)

12. UsukiDoll

range is going to be a little bit harder though...

13. anonymous

Is it all real numbers except y=1?

14. UsukiDoll

I gotta be honest. I have to graph that function first.. and desmos hates me and my fractions

15. Michele_Laino

I agree with the answer of @UsukiDoll

16. UsukiDoll

thanks, HEY YOUR QH symbol is back @Michele_Laino !

17. Michele_Laino

yes! I see nevertheless not in mathematics subsections, thanks! @UsukiDoll

18. UsukiDoll

19. UsukiDoll

the line has to cover the y-values ... if the line doesn't cover the y-values then we have a restriction. This is where I fumble a lot. It could be all reals except y = 1... not sure. What do you think @Michele_Laino

20. Michele_Laino

at points where the function is undefined, generally that function behaves as an infinity, and that is what is happening into your graph

21. UsukiDoll

so it's all reals for the range?

22. UsukiDoll

it's like every time I do these problems, I understand the domain no problem. I see the gaps. It's the range that's frustrating.

23. Michele_Laino

the range, as in this case, is always a real set, nevertheless is not a limited set

24. UsukiDoll

because it goes on forever?

25. Michele_Laino

yes!

26. UsukiDoll

|dw:1437990839458:dw| ok.. so consider this drawing of a circle .. would the domain be from -1 to 1 and then range be -1 to 1 as well.

27. Michele_Laino

towards down or towards up

28. arindameducationusc

yes, range should be -1 to 1

29. arindameducationusc

y-axis represents Range

30. Michele_Laino

your function is a circular function, so its domain is the subsequent set: $\Large [0,2\pi )$

31. UsukiDoll

*I'm not going for trig... just regular domain and range*

32. Michele_Laino

in order to establish the range, we need to know the exact algebraic shape of your function

33. arindameducationusc

One second... A circle is not a function... right?

34. arindameducationusc

As intersects two times along x-axis.

35. UsukiDoll

it's not a function.. fails the vertical line test

36. Empty

Depends on what coordinate system you're using.

37. UsukiDoll

not 1 to 1 either.. fails horizontal line test.

38. Michele_Laino

another possible case is that of your graph is related to this equation: $\Large {x^2} + {y^2} = 1$ nevertheless as we can easily check, that is not a function

39. arindameducationusc

So, how come range will be there. Not possible

40. Michele_Laino

there is not meaning in speaking about range, since we have not a function

41. Empty

$$r(\theta) = 1$$ is a function for the domain $$[0,2 \pi)$$ that is a circle. Vertical line test is for boxes not circles :P

42. Michele_Laino

here is why we hav not a function: |dw:1437991311333:dw|

43. UsukiDoll

actually the vertical line test can be used for circles. and yeah I know that.

44. ParthKohli

Guys, I have a neat way to find the range...

45. arindameducationusc

ok, try finding range of inverse of sin(sqrt(x^2+x+1))

46. ParthKohli

$y = \frac{x^2 + x - 2}{x^2 - 3x - 4}$$\Rightarrow yx^2 - 3y x - 4y = x^2 + x - 2$$\Rightarrow (y-1)x^2 -(3y+1)x - (4y+2) = 0$We want to check the values of $$y$$ for which this equation has a real root, so$D \ge 0$$(3y+1)^2 + 4(4y-2)(y-1) \ge 0$$\Rightarrow 9y^2 + 6y + 1 + 4(4y^2 - 2y - 4y + 2) \ge 0$$\Rightarrow 25y^2 -18y + 9 \ge 0$This inequality is always true, so $$y \in \mathbb R$$.

47. ParthKohli

Oh, and $$3/5, 1$$ won't be included.

48. UsukiDoll

oh cool @ParthKohli

49. UsukiDoll

@ParthKohli one question how did you get $(3y+1)^2 + 4(4y-2)(y-1) \ge 0$

50. Michele_Laino

since we have: $\Large {x^2} + x + 1 = {\left( {\sin y} \right)^2}$ the range is the subsequent set: $\Large \begin{gathered} range = \left\{ {x \in {\mathbf{R}}\;{\text{such that:}}\;0 \leqslant {x^2} + x + 1 \leqslant 1} \right\} = \hfill \\ \hfill \\ = \left\{ {x \in {\mathbf{R}}\;{\text{such that:}}\; - 1 \leqslant x \leqslant 0} \right\} \hfill \\ \end{gathered}$

51. ParthKohli

OK, let me try that out again... we want to check the values of $$y$$ where the quadratic equation$(y-1)x^2 - (3y + 1) x - (4y - 2) = 0$has real solutions.$\rm discriminant \ge 0 \tag{for real solutions}$

52. Michele_Laino

whereas the domain, is: $\Large domain = \left\{ {\forall y \in {\mathbf{R}}} \right\}$

53. ParthKohli

...yes.

54. UsukiDoll

OH THAT IS COOL! I AM SO USING THAT! Thanks @ParthKohli

55. ParthKohli

:)

56. arindameducationusc

Thank you @Michele_Laino so in terms of pi? But in my book the answer is [pi/3,pi/2]

57. Michele_Laino

I'm pondering...

58. arindameducationusc

range | sin^(-1)(sqrt(x^2+x+1))

59. UsukiDoll

in degree mode [60 degrees , 90 degrees] hmmm...

60. arindameducationusc

I searched wolfram... answer is coming.. but if any of you guys can give an explanation it would be awesome

61. Michele_Laino

sorry my answer above is an error, here is the right answer: $\Large {x^2} + x + 1 = {\left( {\arcsin y} \right)^2}$

62. Michele_Laino

range is such that: $\Large 0 \leqslant \sqrt {{x^2} + x + 1}$

63. Michele_Laino

we have this: $\Large {x^2} + x + 1 \geqslant \frac{3}{4}$

64. Michele_Laino

we can try to solve this condition: $\Large \frac{{\sqrt 3 }}{2} \leqslant \sqrt {{x^2} + x + 1} \leqslant 2\pi$

65. arindameducationusc

hmmm... analyzing

66. arindameducationusc

nope, I am not getting. I will think more...