- Messages
- 12
- Reaction score
- 5
- Points
- 13
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
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
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
Can u explain the graph part? How we put the equality signGuys tell me wat u dun get? panda @syed1996
U kno 2 values of k but dunno wat in equality to put ryt?
this question was solved a while ago in this thread i think ...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) .
ExactlyGuys tell me wat u dun get? panda @syed1996
U kno 2 values of k but dunno wat in equality to put ryt?
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.
yeah i dont get why they donot search the threadthis question was solved a while ago in this thread i think ...
this question was solved a while ago in this thread i think ...
i guess its kinda tedious to go through 400 pagesyeah i dont get why they donot search the thread
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!
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)]
Yes.Thank you both so much! For a divergent sequence, r > 1?
Actually its |r|>= 1Thank you both so much! For a divergent sequence, r > 1?
wrong its fg-1 not f-1gIf 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
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!!
it doesnt matter about the modulous in p1 it does in p2 and p3Actually its |r|>= 1
Well, if you learn the right thing from the beginning, wouldnt it be more appropriate?it doesnt matter about the modulous in p1 it does in p2 and p3
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now