Solutions

[tazza]

err.. with the maximal domain.. doesnt it have to be defined for both f(x) and g(x)??

cause between -2 and 3 is not defined for f(x)

Yeah but you're putting that x value into g(x) first, and then the answer for g(x) is then the x value that gets 'processed' by the f(x) function

[Max Peters]

But that cancel's out when you put in values for f(g(x)). I think.

[knightstarr17]

What was Q2(b) asking again?

can someone please answer this question

[extcar]

4^x - 15 * 2^x = 16 or something

[b^3]

4^x-15*2^x=16 or something like that. The answer was x=4.

EDIT: beaten by 9 seconds due to using supscript.

[knightstarr17]

Thank you guys

[Kingofrok33]

How the hell were we meant to do that? If x=4 then the whole thing is -3824, which definitely doesn't look like 16.

[skorm123]

Anyone wanna put up there answers?
I got 19+3√5 for the max length.
i got a = 8 and m = 4 and
the domain question (-infinity,-8] U [-2,infinity)
thats all i can really remember

yes i got the same as you for both, i used pythag for the last equation but im not sure for the domain

[Andiio]

How the hell were we meant to do that? If x=4 then the whole thing is -3824, which definitely doesn't look like 16.

4^4 - 15*(2^4) = 16

[skorm123]

How the hell were we meant to do that? If x=4 then the whole thing is -3824, which definitely doesn't look like 16.

4^4 - 15*(2^4) = 16

let a variable say t= 2^x and then use the quadratic eqn

[t5am94]

Fuck! I mad too many stupid mistakes. Looks like its gonna be 25-30/40 for me

[Kingofrok33]

Oh right, I put the numbers the wrong way around 

[Kingofrok33]

4^x - 15*(2^x) = 16

2^x = a

2a - 15a =16, a=-16/13

Therefore, x=logbase(2)(-16/13)

This is what I did, and I have no idea how I'm meant to get 4 from this.

[vgardiy]

Kingo, in your third line that should be a^2, not 2a

[b^3]

4^x - 15*(2^x) = 16

2^x = a

2a - 15a =16, a=-16/13

Therefore, x=logbase(2)(-16/13)

This is what I did, and I have no idea how I'm meant to get 4 from this.

a^2-15a-16=0
(a-16)(a+1)=0
a=16 or a=-1
2^x=16
2^x=2^4
x=4
2^x=-1, no solution.