g(x) = 3x - 4
f(g(x)) = 9x² - 6x + 1
Find f(x). I found by trial and error, meaning, making a trial and compensating for the errors that didn't match the the composite function, but I'd like to know if there's a procedure to get it. Thanks.
f(g(x)) = 9x² - 6x + 1
Find f(x). I found by trial and error, meaning, making a trial and compensating for the errors that didn't match the the composite function, but I'd like to know if there's a procedure to get it. Thanks.
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What I would do:
We have f(g(x))= 9x² - 6x + 1
Wherever you see g(x), write 3x-4, since g(x) = 3x - 4
f(g(x))= 9x² - 6x + 1
f(3x-4) = 9(3x-4)² -6(3x-4) + 1
Now, since you're looking for f(x), you have to find a value that if you put in 3x-4, it would make x, which would be (x/3) + (4/3) or (x+4)/3
Checking: 3( x/3 + 4/3) -4 -> x +4 - 4 -> x It works!
So now, wherever you see x, write (x+4)/3
f(3x-4) = 9(3x-4)² -6(3x-4) + 1
f( 3((x+4)/3) - 4) = 9( 3((x+4)/3) -4)² - 6 (3 ((x+4)/3) -4) + 1
f(x) = 9( 3((x+4)/3) -4)² - 6 (3 ((x+4)/3) -4) + 1
If you basicallly solve the long equation, you'll get f(x)!
Hope you understand this, even though it looks a bit confusing when you first look at it!
We have f(g(x))= 9x² - 6x + 1
Wherever you see g(x), write 3x-4, since g(x) = 3x - 4
f(g(x))= 9x² - 6x + 1
f(3x-4) = 9(3x-4)² -6(3x-4) + 1
Now, since you're looking for f(x), you have to find a value that if you put in 3x-4, it would make x, which would be (x/3) + (4/3) or (x+4)/3
Checking: 3( x/3 + 4/3) -4 -> x +4 - 4 -> x It works!
So now, wherever you see x, write (x+4)/3
f(3x-4) = 9(3x-4)² -6(3x-4) + 1
f( 3((x+4)/3) - 4) = 9( 3((x+4)/3) -4)² - 6 (3 ((x+4)/3) -4) + 1
f(x) = 9( 3((x+4)/3) -4)² - 6 (3 ((x+4)/3) -4) + 1
If you basicallly solve the long equation, you'll get f(x)!
Hope you understand this, even though it looks a bit confusing when you first look at it!
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Somehow we have to get from 3x to 9x^2. We can do this by squaring g(x)
Suppose h(x) = x^2. Then h(g(x)) = (3x - 4)^2 = 9x^2 - 24x + 16
If we subtract
f(g(x)) - h(g(x)) = (9x^2 - 6x + 1) - (9x^2 - 24x + 16)
-6x + 1 + 24x - 16
18x - 15
Now we need to get to 18x from 3x. We can do this by multiplying by 6.
h(x) = x^2 + 6x
h(g(x)) = (3x - 4)^2 + 6 * (3x - 4) = 9x^2 - 24x + 16 + 18x - 24 = 9x^2 - 6x - 8
f(g(x)) - h(g(x)) = (9x^2 - 6x + 1) - (9x^2 - 6x - 8)
= 1 + 8 = 9
So if h(x) = x^2 + 6x + 9, we should get the right answer.
g(0) = -4, h(-4) = 1, f(g(0)) = 1
g(1) = -1, h(-1) = 4, f(g(1)) = 4
g(2) = 2, h(2) = 25, f(g(2)) = 25
g(-3) = -13, h(-13) = 100, f(g(-3)) = 100
So you can see that my h(x) matches the desired f(x).
Suppose h(x) = x^2. Then h(g(x)) = (3x - 4)^2 = 9x^2 - 24x + 16
If we subtract
f(g(x)) - h(g(x)) = (9x^2 - 6x + 1) - (9x^2 - 24x + 16)
-6x + 1 + 24x - 16
18x - 15
Now we need to get to 18x from 3x. We can do this by multiplying by 6.
h(x) = x^2 + 6x
h(g(x)) = (3x - 4)^2 + 6 * (3x - 4) = 9x^2 - 24x + 16 + 18x - 24 = 9x^2 - 6x - 8
f(g(x)) - h(g(x)) = (9x^2 - 6x + 1) - (9x^2 - 6x - 8)
= 1 + 8 = 9
So if h(x) = x^2 + 6x + 9, we should get the right answer.
g(0) = -4, h(-4) = 1, f(g(0)) = 1
g(1) = -1, h(-1) = 4, f(g(1)) = 4
g(2) = 2, h(2) = 25, f(g(2)) = 25
g(-3) = -13, h(-13) = 100, f(g(-3)) = 100
So you can see that my h(x) matches the desired f(x).
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f(g(x)) = f(3x - 4) = 9*x^2 - 6x + 1.
Assume f(y) = a*y^2 + b*y + c. Then
f(3x - 4) = a*(3x - 4)^2 + b*(3x - 4) + c =
a*(9*x^2 - 24x + 16) + b*(3x - 4) + c =
9a*x^2 + (-24a + 3b)*x + (16a - 4b + c). So for this to equal 9*x^2 - 6x + 1 for all x
we need to coefficients to match, i.e., we require that
9a = 9, -24a + 3b = -6 and 16a - 4b + c = 1. Now
9a = 9 ----> a = 1, and so
-24a + 3b = -24*1 + 3b = - 6 ----> 3b = 18 ----> b = 6, and finally
16a - 4b + c = 16*1 - 4*6 + c = 1 ----> c = 9.
Thus f(x) = x^2 + 6x + 9.
So if we assume that f(x) has the form of a quadratic polynomial, f(x) must
be x^2 + 6x + 9. This does not prove that this solution is unique, but I'm guessing
that this solution is all you were looking for. :)
Assume f(y) = a*y^2 + b*y + c. Then
f(3x - 4) = a*(3x - 4)^2 + b*(3x - 4) + c =
a*(9*x^2 - 24x + 16) + b*(3x - 4) + c =
9a*x^2 + (-24a + 3b)*x + (16a - 4b + c). So for this to equal 9*x^2 - 6x + 1 for all x
we need to coefficients to match, i.e., we require that
9a = 9, -24a + 3b = -6 and 16a - 4b + c = 1. Now
9a = 9 ----> a = 1, and so
-24a + 3b = -24*1 + 3b = - 6 ----> 3b = 18 ----> b = 6, and finally
16a - 4b + c = 16*1 - 4*6 + c = 1 ----> c = 9.
Thus f(x) = x^2 + 6x + 9.
So if we assume that f(x) has the form of a quadratic polynomial, f(x) must
be x^2 + 6x + 9. This does not prove that this solution is unique, but I'm guessing
that this solution is all you were looking for. :)