If (f o g)= 16x + 7 and g(x)= 2x - 1 , find f(x)

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To find f(x) given (f o g) = 16x + 7 and g(x) = 2x - 1, substitute g(x) into the composite function. The correct approach involves assuming f(x) is a linear function of the form Ax + B. By substituting g(x) into f(g(x)), the goal is to equate it to 16x + 7 to solve for A and B. The discussion also suggests verifying the result by checking if f(g(x)) returns the original expression. The conversation highlights the importance of methodical substitution in function composition.
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Homework Statement



If (f o g)= 16x + 7 and g(x)= 2x - 1 , find f(x)


The Attempt at a Solution



I haven't done a problem like this in awhile so I don't remember exactly what to do. I think you plug g(x) into (f o g) to get 32x - 9 but I'm not sure if that's correct.

Homework Statement

 
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Andolph23 said:

Homework Statement



If (f o g)= 16x + 7 and g(x)= 2x - 1 , find f(x)


The Attempt at a Solution



I haven't done a problem like this in awhile so I don't remember exactly what to do. I think you plug g(x) into (f o g) to get 32x - 9 but I'm not sure if that's correct.

Homework Statement


If this is correct, you should try f(g(x)) and see if you get what it was originally

P.S. Notice your trying to get f(x)
 
Suppose that f(x) = Ax + B, a linear function. Given that, and that g(x) = 2x - 1, what is (f o g)(x)?
(Ignore for a moment that (f o g)(x) was already given.)

P.S. This should really be in the Precalculus subforum.
 
What is (f○g○g-1)(x) ?
 
SammyS said:
What is (f○g○g-1)(x) ?

Wow, your good:wink: Didn't even think of that.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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