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## anonymous 5 years ago I've got a math exam on friday. I have some of the previos exams here and there are a whole bunch of questions which I cant solve. I really needs some help with them. I'll be posting them here now. Thank you very much for all who help me out, I really appreciate it :)

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1. anonymous

Limits problems, find the limit: a) $\lim_{x \rightarrow 5}x/(x-5)$ b) $\lim_{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})$

2. anonymous

a) add and subtract 5 frm d numerator that gives u 1 + 1/(x-5) when u put in infinity it is 1 + 1/inf = 1

3. anonymous

as for b) multiply and divide by $\sqrt{x+1}+\sqrt{x}$

4. anonymous

that gives u 1 / √x+1+√x

5. anonymous

when u put in infinity it gives u 1/inf = 0

6. anonymous

$a) \lim_{x \rightarrow 5} { x \over x-5} =\infty$ That's because when you substitute 5 in the top you get 5, and in the bottom you get 0. A number over 0 is infinity.

7. anonymous

Next up, find the derivative: a) $f(x) =x^{2}\times e^{x ^{2}}$ b) $f(x,y)=(y ^{3}x+e ^{2y})^{3}$

8. anonymous

a) dy/dx= 2xe^x3 + 3x^4e^x^3

9. anonymous

b) him1618 already got the right answer.

10. anonymous

was a wrong?

11. anonymous

I guess :)

12. anonymous

By the way, in a) if the limit approaches 5 from the left, it would be -infinity.

13. anonymous

so ur saying it doesnt exist?

14. anonymous

Yep.

15. anonymous

yeah neway its a sort of hyperbola centred at 5

16. anonymous

It has a vertical asymptote at x=5.

17. anonymous

Question 3: Define the min/max of $f(x)=x^{3}-12x ^{2}+21x+25$ in [-1, infinity[

18. anonymous

differentiate it f'(x) = 3x^2 - 24x + 21

19. anonymous

the roots are 7, 1

20. anonymous

Question 4: Define the local min/max of $f(x,y)=x ^{3}-y ^{3}+3xy$

21. anonymous

Sick how good you are :) Two more questions...

22. anonymous

yeah go on

23. anonymous

Question 5: Solve the following inequalities a) $x^{3}+2x ^{2}+2x \le0$ b) $\log_{2} x <2+\log_{4} x$

24. anonymous

Question 6: Solve the following equations a) $\left| x-3 \right|+\left| x+2 \right|=10$ b) $2^{3x}=3^{2x}$

25. anonymous

a) x<=0

26. anonymous

is it right

27. anonymous

I believe it's right since it's factored to x(x^2+2x+2). But the part x^2+2x+2 is always >0. So, the sign just depends on x. Hence the whole expression is <=0 when x<=0.

28. myininaya

anwar is like totatally right!

29. anonymous

so am i man..dont demean me..lol

30. anonymous

Why is the part x^2+2x+2 always >0?

31. anonymous

because its determinant is negative...

32. anonymous

Yeah. You're the one who got the answer. I was just following your foot steps :)

33. anonymous

it is an upward parabola where the min value is greater than 0

34. anonymous

Ah I understand :)

35. anonymous

You deserve a medal him1618.

36. anonymous

thnx a lot

37. anonymous

I gave one too :)

38. anonymous

You're welcome :)

39. anonymous

so kind

40. anonymous

I want to solve part b) in inequalities.

41. anonymous

b) x is in 0 to 16

42. anonymous

Please do I can't solve it :/

43. anonymous

thats d ans

44. anonymous

its damn tedious to write these eqns

45. anonymous

Yeah, that sucks....

46. anonymous

on d comp only..on paper it took a minute

47. myininaya

i think i considered all the cases for 6a

48. anonymous

2 is $\log_{4}16$

49. anonymous

$\log_2x-\log_4x<2 \implies \log_2x-{\log_2x \over \log_24}<2 \implies \log_2x-{1 \over 2}\log_2x<2$ ${1 \over 2}\log_2 x<2 \implies \log_2{\sqrt x}<2 \implies \sqrt x<4 \implies x<16$

50. anonymous

But x has to be a positive number, So 0<x<16

51. anonymous

right

52. anonymous

Nice answer myininaya helpd me a bunch :)

53. myininaya

here is 6b

54. anonymous

yeah 6a is bloody tedious..nice wrk

55. anonymous

b is right 2

56. anonymous

Right on the spot AnwarA I'll jot that down to my notebook :)

57. anonymous

Did anyone solve 6a, or I can take it?

58. anonymous

yes its been solved by myininaya

59. anonymous

I hate you man >.<

60. myininaya

woman* lol

61. anonymous

OMG.. I don't hate you at all :P

62. myininaya

lol

63. myininaya

you can look over my work. there might be a mistake who knows

64. myininaya

it looks good to me though

65. anonymous

I can hardly read it :P

66. myininaya

:(

67. myininaya

i can rewrite it

68. anonymous

No, It's right. I got the exact same answers.

69. myininaya

you can also graph it in your calculator if you want to verify your answers

70. anonymous

x=15/2 for x>=3, x=-5/2 for x<=2. No solution for 2<x<3.

71. myininaya

right!

72. myininaya

i can write neat when i'm not trying to beat you guys to answer the questions lol

73. myininaya

i wish in high school there was a math quiz bowl and i participated that would have been so fun

74. anonymous

lol we should divide the questions equally. I wish I had a scanner -.-

75. anonymous

One question, what happens here: $\log_2x-{\log_2x \over \log_24}<2 \implies \log_2x-{1 \over 2}\log_2x<2$ ?

76. anonymous

log4 to base to is 2.

77. anonymous

to base 2*

78. myininaya

because 2^2=4

79. anonymous

Mm, yes of course! :)

80. anonymous

Do you still have any questions, or that's it?

81. anonymous

I think that is about it :) So quick and dirty. Thanks a million for all of you!

82. myininaya

yes all your questions were being answered so quickly by him hes such a meanie

83. anonymous

no probs mate

84. anonymous

yeah thats kind...

85. myininaya

lol i'm kidding

86. anonymous

dont sweat..i hate studying so much...bt i lov doing this

87. anonymous

I think that is about it :) So quick and dirty. Thanks a million for all of you!

88. anonymous

well thnx

89. anonymous

Good luck in your exam on Friday :)

90. anonymous

yeah good luck...remember me if u feel nervy..lol

91. anonymous

Thanks, I believe I can make it :D

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