Mathematics: Post your doubts here!

[doctorofelectronics]

If f(x)=x²-4x+7 for x>2 and g(x)= x-2 for x>2

The function h is such that f=hg and the domain of h is x>0
Q. Obtain an expression for h(x) .

h(x)=x^2+3

 

[ZainH]

h(x)=x^2+3

How? Explain please =)

 

[Manobilly]

Guys tell me wat u dun get? panda @syed1996
U kno 2 values of k but dunno wat in equality to put ryt?

Can u explain the graph part? How we put the equality sign

 

[daredevil]

If f(x)=x²-4x+7 for x>2 and g(x)= x-2 for x>2

The function h is such that f=hg and the domain of h is x>0
Q. Obtain an expression for h(x) .

this question was solved a while ago in this thread i think ...

 

[PANDA-]

Guys tell me wat u dun get? panda @syed1996
U kno 2 values of k but dunno wat in equality to put ryt?

Exactly

 

[asd]

Somebody please do this!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
Q7 b)

Well for part one you only have to assume that the common ratio r is greater that 0 and less than 1. That goes for a convergent progression.
so your r is= 1/3tan^2(theta)
0<1/3tan^2(theta)<1u'd get the answer.
for (ii) Sum to infinity= a/(1-r)
-> 1/[1-1/3tan^2(pi/6)]

 

[A star]

this question was solved a while ago in this thread i think ...

yeah i dont get why they donot search the thread

 

[haha101]

someone help mee

 

[doctorofelectronics]

Its easy. just imagine in which function of x would you substitute x-2 to get x^2-4x+7
here you need to square the x in x-2 because it is x squared in f(x)
So you square the bracket (x-2) ie: (x-2)(x-2)
You get x^2-4x+4
Now to get f(x) just add 3 to the equation!

 

[ZainH]

this question was solved a while ago in this thread i think ...

If you could link me to the post, or tell me the page number that'd be great =D
If you can't , then could you let me know how to solve it?

 

[daredevil]

yeah i dont get why they donot search the thread

i guess its kinda tedious to go through 400 pages
and we do like half a century of pages in a day here so let's cut them some slack...
i was just making sure it was the same question and i hadn't confused it with another one

 

[ZainH]

Its easy. just imagine in which function of x would you substitute x-2 to get x^2-4x+7
here you need to square the x in x-2 because it is x squared in f(x)
So you square the bracket (x-2) ie: (x-2)(x-2)
You get x^2-4x+4
Now to get f(x) just add 3 to the equation!

I don't get it? ._________.

Since g(x)= x-2 , and f=hg..

if h(x)= x²+3 , won't f=(x²+3)(x-2) ?

And that gives something completely different D:

 

[Magenta]

You should remember that |r| < 1 for a convergant sequence
In this question r = 1/3 tan^2 θ
therefore, 1/3 tan^2 θ < 1
tan^2 θ < 3
tan θ < [sqrt]3
θ < π/3

Well for part one you only have to assume that the common ratio r is greater that 0 and less than 1. That goes for a convergent progression.
so your r is= 1/3tan^2(theta)
0<1/3tan^2(theta)<1u'd get the answer.
for (ii) Sum to infinity= a/(1-r)
-> 1/[1-1/3tan^2(pi/6)]

Thank you both so much! For a divergent sequence, r > 1?

 

[daredevil]

If f(x)=x²-4x+7 for x>2 and g(x)= x-2 for x>2

The function h is such that f=hg and the domain of h is x>0
Q. Obtain an expression for h(x) .


well u know how we to the fg(x) we take the f function and put the g function in the place of x of the f funtion and solve it??
here it is kinda the same deal... the difference is that they have given u another letter to confuse u
if f=hg then f-1(g)=h
now find f^-1 and substitute the g function there in place of x and solve it
what u get for the answer as an expression is h
if u stil don't get it think over it a lott!!! and then take a break and come here again at least some of us will still be here in the wee hours too we're that paranoid

 

[asd]

Thank you both so much! For a divergent sequence, r > 1?

Yes.

 

[asd]

Thank you both so much! For a divergent sequence, r > 1?

Actually its |r|>= 1

 

[A star]

If f(x)=x²-4x+7 for x>2 and g(x)= x-2 for x>2

The function h is such that f=hg and the domain of h is x>0
Q. Obtain an expression for h(x) .


well u know how we to the fg(x) we take the f function and put the g function in the place of x of the f funtion and solve it??
here it is kinda the same deal... the difference is that they have given u another letter to confuse u
if f=hg then f-1(g)=h
now find f^-1 and substitute the g function there in place of x and solve it
what u get for the answer as an expression is h
if u stil don't get it think over it a lott!!! and then take a break and come here again at least some of us will still be here in the wee hours too we're that paranoid

wrong its fg-1 not f-1g

 

[gary221]

Rutzaba... sorry 4 disturbing... ... bt i need help!!!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_ms_3.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_3.pdf

ques 1... pls!!
thnx!!

guys pls!

 

[A star]

Actually its |r|>= 1

it doesnt matter about the modulous in p1 it does in p2 and p3

 

[asd]

it doesnt matter about the modulous in p1 it does in p2 and p3

Well, if you learn the right thing from the beginning, wouldnt it be more appropriate?