Composition of Functions: Finding g(f-1(x)) with Given Equations

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To find g(f-1(x)) for the functions f(x) = 1/3x + 2 and g(x) = 2x² - x + 7, the inverse function f-1(x) is calculated as f-1(x) = 3x - 6. Substituting f-1(x) into g(x) leads to the expression g(3x - 6) = 2(3x - 6)² + (3x - 6) + 7. After simplifying, the result is 18x² - 69x + 73, correcting a previous miscalculation that yielded 71. The discussion highlights the importance of careful algebraic manipulation to avoid sign errors in the final answer.
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Homework Statement



Given: f(x)= 1/3x + 2 and g(x)= 2x2- x + 7

Find: g(f-1(x))

Homework Equations


The Attempt at a Solution



x=1/3y + 2
1/3y= x - 2
y=(x-2)/(1/3)
y=3x-6= f-1(x)

2(3x-6)2 + (3x-6) +7
solved to get..
18x2-69x+71
 
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I got a different answer but your method is correct. You have a sign error somewhere.
 


2(36)- 6+ 7= 72+1= 73, not 71.
 


g(x)= 2x² - x + 7

2(3x-6)² + (3x-6) + 7
 

Thread 'Greatest possible value of a constant in polynomial'

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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