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Composite Functions

f(x)=x2−2x+3[br]g(x)=zx2+1 f(x) = x^2 - 2x + 3[br] g(x) = zx^2 + 1 where z is a constant, x ∈ and real.
Given that gf(2)=16gf(2) = 16 find the value of z.

I am struggling on finding the composite function of gf(x)gf(x) in general. It has been a while since I did composite functions and I have tried with the help of my textbook but I have not succeeded.
Reply 1
Avatar for nuodai
nuodai
17
Well gf(x)gf(x) just means g(f(x))g(f(x)). That is, wherever you see xx in the expression for g(x)g(x), replace it with f(x)f(x).
Reply 2
Avatar for Jimny
Jimny
You don't need the general g(f(x)) You can work out f(2) then use this in g(f(2)) to find your answer.
Reply 3
Avatar for JohnMgee
JohnMgee
OP
Original post by nuodai
Well gf(x)gf(x) just means g(f(x))g(f(x)). That is, wherever you see xx in the expression for g(x)g(x), replace it with f(x)f(x).

So z(x2−2x+3)2+1z(x^2 - 2x + 3)^2 +1?
put x =2 in the above piece. set it equal to 16 and then solve for z.
Reply 5
Avatar for nuodai
nuodai
17
Original post by JohnMgee
So z(x2−2x+3)2+1z(x^2 - 2x + 3)^2 +1?


Yup. Now just plug x=2x=2 into this, evaluate it (in terms of zz), set it equal to 16 and then solve for zz.
Reply 6
Avatar for JohnMgee
JohnMgee
OP
9z+1=169z + 1 = 16 therefore z=15/9z = 15/9 But the answer is 5?

This is where I got to before I made this thread so is it at this part I am going wrong?
Reply 7
Avatar for nuodai
nuodai
17
Original post by JohnMgee
9z+1=169z + 1 = 16 therefore z=15/9z = 15/9 But the answer is 5?

This is where I got to before I made this thread so is it at this part I am going wrong?


The answer's definitely 159\dfrac{15}{9}. The book must be wrong.
Reply 8
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JohnMgee
OP
My book has been known to have many mistakes so it isn't surprising.

Thanks for the help.
Reply 9
Avatar for noobynoo
noobynoo
6
if g(x)=zx+1 then the answer would be 5.
Reply 10
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JohnMgee
OP
It defiantly says g(x)=zx^2+1 so I'd say the mistake was there. well spotted!

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