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Mathematics: Post your doubts here!

ZainH

ZainH

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daredevil said:
this question was solved a while ago in this thread i think ... o_O
Click to expand...

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

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A star said:
yeah i dont get why they donot search the thread :confused:
Click to expand...
i guess its kinda tedious to go through 400 pages :p
and we do like half a century of pages in a day here so let's cut them some slack... :p
i was just making sure it was the same question and i hadn't confused it with another one :p
 
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ZainH

ZainH

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doctorofelectronics said:
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!
Click to expand...

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:
 
M

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Maz said:
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
Click to expand...
asd said:
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)]
Click to expand...

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

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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 :p
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 :p we're that paranoid :p
 
A

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daredevil said:
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 :p
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 :p we're that paranoid :p
Click to expand...
wrong its fg-1 not f-1g
 
ZainH

ZainH

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daredevil said:
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 :p
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 :p we're that paranoid :p
Click to expand...

I actually understood, not from what you explained.. but while reading I realized what we had to do xD
Anyways thanks! You too doctorofelectronics
 
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