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unit 1 - non calculater revision - composite functions?


7 replies to this topic

#1 xClairex

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Posted 15 January 2007 - 07:58 PM

the functions are defined by f'(x)=2x + 5 and g(x)=x - 3

the function h is such that whenever f(p)=q and g(q)=r then h(p)=r

a)if q=7, find the values of p and r

b)find a formula for h(x), in terms of x.

any help would be much appreciated i think i was either off or talking when we got taught this as i can't remember it unsure.gif


#2 Dave

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Posted 16 January 2007 - 01:06 AM

a is simple enough just sub in q=7 and work though all the info you know and you should get p=1 and r= -2 but its possible i am wrong i usually am these days

b not sure about atm

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#3 John

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Posted 16 January 2007 - 11:46 AM

for b Integrate f'(x) and h(x) will either be f(g(x)) or g(f(x))

#4 Dave

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Posted 16 January 2007 - 02:05 PM

meh meant to come back to this but i erm forgot

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#5 xClairex

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Posted 16 January 2007 - 05:21 PM

I know i probably seem really thick but i just dont get it

f'(x)=2x + 5
2x + 5 = 7
2x = 7 - 5
2x = 2
x= 2/2
x=1 so i get p = 1

but i dont understand how you got r = -2

g(x) = x - 3
x - 3 = 7
x= 7 + 3
x=10 i know thats not right how do i get r = -2 ?



#6 Dave

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Posted 16 January 2007 - 06:21 PM

sorry i read the question wrongly i read g(p) not g(q)

anyway

g(7)=49-3 = r

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#7 xClairex

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Posted 16 January 2007 - 06:25 PM

its fine so r = 46?


#8 xClairex

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Posted 17 January 2007 - 04:43 PM

i now know how to do this thanx for the help






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