Solving problems in function notation

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The discussion revolves around solving for g(3) using the function notation provided. Given that g(1) = 3 and the equation g(2x + 1) = 4g(x) + x + 1, participants explore how to derive g(3). By substituting x = 1 into the equation, they find that g(3) can be calculated as g(3) = 4g(1) + 2, leading to the solution g(3) = 14. The approach emphasizes understanding the transformations of the function and utilizing known values effectively. The final answer confirms that g(3) equals 14.
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Homework Statement


Given that:
g (1)=3
g (2x+1)=4g (x)+x+1

Find g (3)

Homework Equations


The awnser is 14. (Taken out of a PSAT workbook)

The Attempt at a Solution


I assume that g(2x+1) refers to a series of transformations (horizontal dilation, up 1) of g(x)
Once g(x) is found, g(3) can easily be solved. The problem is, that I have no idea how to find it.
 
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The second equation tells you how to find g(2x + 1) if you already know g(x).

For which value of x do you already know g(x)?
 
Daringpear said:

Homework Statement


Given that:
g (1)=3
g (2x+1)=4g (x)+x+1

Find g (3)

Homework Equations


The awnser is 14. (Taken out of a PSAT workbook)

The Attempt at a Solution


I assume that g(2x+1) refers to a series of transformations (horizontal dilation, up 1) of g(x)
Once g(x) is found, g(3) can easily be solved. The problem is, that I have no idea how to find it.
To me it appears you think a little bit too complicated.
Can you write ##3## as ##2x+1## and what does it tell you about ##x##?
 
Ah! Your suggestions make sense!

When substituting x for one,

g(2(1)+1)=4g(1)+1+1
g(3)=4(3)+2
g(3)=12+2
g(3)=14
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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