functions help Watch

panjabiflower
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#1
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#1
1. the functions f and g are defined by
f:x--> 5^x -7 xER g:x--> 2x + 3 xER
a) find and simplify an expression for gf stating its domain
b) solve gf(x)=10

2) f:x--> e^1/2x -2 xER
a) evaluate f(ln9)
b) state range of f

3) f(x)= x^2 -4x+5 xER x> or equal to 2
a) f(x) in form a(x+b)^2 + c
b) state range of f

4) f:x--> x^2+4 xER
g:x--> 2x -1/x xER
a) what is gf(-2)
b) expression for fg(x)
values of x for which fg(x)=5
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RichE
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Do you know what the composition gf means? If so then what's the problem? Can we see some working?
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panjabiflower
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if you are specifically referring to 1 B) then when i worked out i got the answer to be 7.85 what did you get??
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DFranklin
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Do you not understand what "Can we see some working?" means?
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panjabiflower
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#5
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um sorry...
ok this is how i worked out the 1st questionpart a) 2(5^x-7)+3
=10^x-7+3
(don't know about the domain)
b) log10^x-7=log7
...
x=log7+7log10/log10=7.85



and i am completely confused with the other questions....
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1721
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#6
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a hint g(f(x)) so f(x) is X in g(x)
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panjabiflower
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#7
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i am still abit stuck
for q1 i got the answer 7.85 is that right?? if not, do you know how to get the correct answer
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1721
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#8
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im bored so i'll give it ago and get back to you
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panjabiflower
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#9
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f(x) is th second one
-sorry about that i didn't make it clear in the question
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RichE
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(Original post by panjabiflower)
um sorry...
ok this is how i worked out the 1st questionpart a) 2(5^x-7)+3
=10^x-7+3
Your algebra is wrong here. You've used a power rule wrong and expanded the bracket wrong.
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1721
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#11
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i get x to be 1.460991592
for 1
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The Muon
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Start by substituting f(x) into g(x) and letting that equal 10. After some cancelling, you should be ablle to take logs of both sides and i think the answer is \frac{log5}{log11.5} but i could have mad a mistake myself
Stil workig on the second bit
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1721
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(Original post by The Muon)
Start by substituting f(x) into g(x) and letting that equal 10. After some cancelling, you should be ablle to take logs of both sides and i think the answer is \frac{log5}{log11.5} but i could have mad a mistake myself
Stil workig on the second bit
you have the fraction upside down.
its xlog5=log(11.5)
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The Muon
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(Original post by 1721)
you have the fraction upside down.
its xlog5=log(11.5)

I knew it looked abit funny - you'd have thought that after a Maths A level I would be able to divide properly
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Kolya
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You seem to have many theoretical problems. Do you not understand the concept of composition of functions at all? If not, I recommend you reread the relevant section in your textbook, including working through the examples, and then come back and ask about any difficulties you have understanding the theory. Question 3a) is just completing the square, surely you know how to do that from your GCSE?! For 1a), You have said that 2(5^x - 7) + 3 = 10^x - 7 + 3 but this is just clumsiness. A little manipulation shows that 2(5^x - 7) + 3 = 2\cdot 5^x - 14 + 3 = 2 \cdot 5^x - 11. You need to remember what a power means. 5^x means 5 \cdot 5 \cdot 5 \cdot... \cdot 5 an x number of times. 10^x is 10 \cdot 10 \cdot 10 \cdot... \cdot 10 an x number of times. How is multiplying the expression for 5^x going to give us the expression for 10^x? If you don't understand how to manipulate algebra then you're handicapped before you start.

Anyway, I recommend the first thing you do is look at the theory and examples in your textbook. There is nothing "out of the ordinary" in the questions you have posted.
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1721
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#16
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(Original post by Kolya)
Anyway, I recommend the first thing you do is look at the theory and examples in your textbook. There is nothing "out of the ordinary" in the questions you have posted.
yeah there not the hardest ive seen.
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