# Thread: reverse composition of functions......:(

1. ## reverse composition of functions......:(

(a) If g(x) = 3x + 2 and h(x) = 9x2 + 12x + 9, find a function f such
that f o g = h. (Think about what operations you would have to perform on the formula for g to end up with the formula for h.)
f = 1
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(b) If f(x) = 3x + 9 and h(x) = 3x2 + 3x + 6, find a function g such
that f o g = h.
g(x) = 2
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how would i do such a question??

2. Originally Posted by Sneaky
(a) If g(x) = 3x + 2 and h(x) = 9x^2 + 12x + 9, find a function f such that f o g = h. (Think about what operations you would have to perform on the formula for g to end up with the formula for h.)

f(x) = x^2 + 5

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(b) If f(x) = 3x + 9 and h(x) = 3x^2 + 3x + 6, find a function g such
that f o g = h.

g(x) = x^2 + x - 1
for future reference, note that exponents can be designated by the caret symbol ...

x^2 is x "squared"

3. can u show me the exact steps into how you got that?

4. Originally Posted by Sneaky
can u show me the exact steps into how you got that?
no formal steps ... I just played with the given functions.

for instance ...

f(3x+2) = 9x^2 + 12x + 9

if (3x+2) is squared, you get 9x^2 + 12x + 4

note that you are short 5 units of the constant value. so ...

f(x) = x^2 + 5

for the second ...

f[g(x)] = 3[g(x)] + 9 = 3x^2 + 3x + 6

note that h(x) has coefficients of 3 for the first two terms ...

so I tried 3(x^2 + x) + 9 = 3x^2 + 3x + 9

but, the constant is 6 ... I have 3 units too many. How can I get rid of the additional 3 ?

3(x^2 + x - 1) + 9 = 3x^2 + 3x - 3 + 9 = 3x^2 + 3x + 6

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### reverse h composition

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