question (1 Viewer)

[Trev]

hey u r guna hate me... but that sine question... i didnt mean to put the "-ve" sign in the question at the start... im sorry.. is it much different from what uve done anywayz?

It will be the same process, just follow it through and see what is different for yourself.

 

[Trev]

x/(x-1)+2/(2x+1)=2
Multiply by (x-1) and (2x+1)
x(2x+1)+2(x-1)=(x-1)(2x+1)
2x²+x+2x-2=2x²+x-2x-1
etc. Solve the quadratic.

 

[auzbabe4christ]

It will be the same process, just follow it through and see what is different for yourself.

yea, thats kool... but ive got to "cos x=(-√2)/2"... that cant solve can it?? or is it just me being blonde an not seeing it??

 

[auzbabe4christ]

x/(x-1)+2/(2x+1)=2
Multiply by (x-1) and (2x+1)
x(2x+1)+2(x-1)=(x-1)(2x+1)
2x²+x+2x-2=2x²+x-2x-1
etc. Solve the quadratic.

ok, i figured out where i went wrong in this thanx heaps.. i did a really stupid mistake. go figure. lol

 

[Trev]

cos x=-(√2)/2
Using 2nd and 3rd quadrants and pi/4 as the basic angle.
x=3pi/4,5pi/4,3pi/4+2pi,5pi/4+2pi...
[Not sure if this next part is completely right but it still works]
x=(180+/-45)+/-360n
The above in degree's, 180 takes you between 2nd and 3rd quadrants, plus/minus 45 (the basic angle) gives you a solutions for x and plus/minus 360n gives another whole revolution around to the same position which will give you the same solution.
x=(180+/-45)+/-360n = (pi+/-pi/4)+/-2pi.n = pi[1 +/- 1/4 +/- 2n].
Hope that's right it seems a bit wierd but I forget the general solutions.

 

[auzbabe4christ]

yea, its alright.. i tried to figure it out an it seemed to work. lol.

oi... with differentiating logs... i know the basic stuff but with this question..

ln(x/(x^2-4)) i cant remember how to do it..

 

[auzbabe4christ]

and whats the basic rule for limits?? i never understood them properly.. cuz i have a question..

lim (x-> -pi/4) cos4x/sin2x. and i cant remember how to do it.. do i just place the "-pi/4" as x?? or what??

 

[auzbabe4christ]

ok.. this is sounding bad cuz i cant do easy stuff now.. but i have a question which yet again isnt falling out for me... ive tried it like 8 times.. it is..

find an expression for: (3/x^2-4)-(x-2)

 

[Trev]

ln(x/(x²-4))
Break it into two logarithms
lnx - ln(x²-4)
d/dx[ln(f(x))] = [f'(x)]/[f(x)]
So:
d/dx[lnx - ln(x²-4)]
= 1/x - 2x/(x²-4) then simplify into one fraction.
Or you can do it from the original question and use the product rule or other rule (dividing one)...

 

[Trev]

lim (x-> -pi/4) cos4x/sin2x. and i cant remember how to do it.. do i just place the "-pi/4" as x?? or what??

Yes, in this instance you can just replace x for -pi/4 because you will not end up with a zero denominator.

 

[auzbabe4christ]

alrighty kool... i did it thru the quotient rule the first time i did it.. so aparentyl i did it right. lol. thanx

 

[auzbabe4christ]

Yes, in this instance you can just replace x for -pi/4 because you will not end up with a zero denominator.

ok, well what about with

lim (x->4) (x-2)/(x^2-16)

and

lim (x->3) (x-3)/(x^2-9)

cuz in both cases u get a 0 on the bottom...

 

[Trev]

3/(x²-4)-(x-2)
[3-(x-2)(x²-4)]/(x²-4)
Expand and simplify.

 

[auzbabe4christ]

alrighty kool... i did it thru the quotient rule the first time i did it.. so aparentyl i did it right. lol. thanx

alright ive come up with two different answers... lol...

mayb b a stupid quesiton... but how am i simplfying the fractions in the differentitating quesiton?? it just isnt working for me...

 

[Trev]

lim (x->3) (x-3)/(x²-9)
lim (x->3) (x-3)/[(x-3)(x+3)]
lim (x->3) 1/(x+3)
= 1/6.
Not sure about the other one.

 

[auzbabe4christ]

lim (x->3) (x-3)/(x²-9)
lim (x->3) (x-3)/[(x-3)(x+3)]
lim (x->3) 1/(x+3)
= 1/6.
Not sure about the other one.

oh ok, i get it for that one. thx

 

[auzbabe4christ]

alright ive had another mental mind blank with this lot of questions... i dont think i learnt this bit properly.. lol..

anywayz..

"Let H be the function given by H(t)= -1, Find:

a) H(0)
b) H(x+h)
c) H(3t)
d H(t+2)-H(t)

 

[Trev]

You're just getting someone to answer all the questions you need done aren't you?
H(t)=-1.
Everything will equal -1 because there are no t's there (variables), it's a straight line and can't be changed.

 

[auzbabe4christ]

You're just getting someone to answer all the questions you need done aren't you?
H(t)=-1.
Everything will equal -1 because there are no t's there (variables), it's a straight line and can't be changed.

no im honestly not... ive serisouly done the rest.. except for a few things which i can do myself (i think) lol.

so for the ones which have "t" in them i am able to work out then?

oh an just cuz i did ext maths doesnt mean i was heaps good at it..

 

[acullen]

lim_(x→4) (x-2)/(x²-16)
=(4-2) / (4²-16)
=2 / (16-16)
=2 / 0
= ∞
Limit Does Not Exist (D.N.E.)

What Maths subject is this for? MATH141/187?