Question

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 :)

Answer 1

Limits problems, find the limit:

a) \[\lim_{x \rightarrow 5}x/(x-5)\]
b) \[\lim_{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})\]

Answer 2

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

Answer 3

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

Answer 4

that gives u 1 / √x+1+√x

Answer 5

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

Answer 6

\[a) \lim_{x \rightarrow 5} { x \over x-5} =\infty\]
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.

Answer 7

Next up, find the derivative:

a) \[f(x) =x^{2}\times e^{x ^{2}}\]
b) \[f(x,y)=(y ^{3}x+e ^{2y})^{3}\]

Answer 8

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

Answer 9

b) him1618 already got the right answer.

Answer 10

was a wrong?

Answer 11

I guess :)

Answer 12

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

Answer 13

so ur saying it doesnt exist?

Answer 14

Yep.

Answer 15

yeah neway its a sort of hyperbola centred at 5

Answer 16

It has a vertical asymptote at x=5.

Answer 17

Question 3:

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

Answer 18

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

Answer 19

the roots are 7, 1

Answer 20

Question 4:

Define the local min/max of
\[f(x,y)=x ^{3}-y ^{3}+3xy\]

Answer 21

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

Answer 22

yeah go on

Answer 23

Question 5:

Solve the following inequalities

a) \[x^{3}+2x ^{2}+2x \le0\]
b) \[\log_{2} x <2+\log_{4} x\]

Answer 24

Question 6:

Solve the following equations

a) \[\left| x-3 \right|+\left| x+2 \right|=10\]
b) \[2^{3x}=3^{2x}\]

Answer 25

a) x<=0

Answer 26

is it right

Answer 27

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.

Answer 28

anwar is like totatally right!

Answer 29

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

Answer 30

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

Answer 31

because its determinant is negative...

Answer 32

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

Answer 33

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

Answer 34

Ah I understand :)

Answer 35

You deserve a medal him1618.

Answer 36

thnx a lot

Answer 37

I gave one too :)

Answer 38

You're welcome :)

Answer 39

so kind

Answer 40

I want to solve part b) in inequalities.

Answer 41

b) x is in 0 to 16

Answer 42

Please do I can't solve it :/

Answer 43

thats d ans

Answer 44

its damn tedious to write these eqns

Answer 45

Yeah, that sucks....

Answer 46

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

Answer 47

i think i considered all the cases for 6a

Answer 48

2 is \[\log_{4}16 \]

Answer 49

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

Answer 50

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

Answer 51

right

Answer 52

Nice answer myininaya helpd me a bunch :)

Answer 53

here is 6b

Answer 54

yeah 6a is bloody tedious..nice wrk

Answer 55

b is right 2

Answer 56

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

Answer 57

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

Answer 58

yes its been solved
by myininaya

Answer 59

I hate you man >.<

Answer 60

woman* lol

Answer 61

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

Answer 62

lol

Answer 63

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

Answer 64

it looks good to me though

Answer 65

I can hardly read it :P

Answer 66

:(

Answer 67

i can rewrite it

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

right!

Answer 72

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

Answer 73

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

Answer 74

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

Answer 75

One question, what happens here:

\[\log_2x-{\log_2x \over \log_24}<2 \implies \log_2x-{1 \over 2}\log_2x<2\]

?

Answer 76

log4 to base to is 2.

Answer 77

to base 2*

Answer 78

because 2^2=4

Answer 79

Mm, yes of course!

:)

Answer 80

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

Answer 81

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

Answer 82

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

Answer 83

no probs mate

Answer 84

yeah thats kind...

Answer 85

lol i'm kidding

Answer 86

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

Answer 87

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

Answer 88

well thnx

Answer 89

Good luck in your exam on Friday :)

Answer 90

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

Answer 91

Thanks, I believe I can make it :D