Problem has me troubled: Let F(x) be a linear function

[rofl]

Let F(x) be a linear function such that for all x, f(3x+3)=4x+1. Find F(x)

I dont even know where to begin...

 

[Mrspi]

Re: Problem has me troubled...

Let F(x) be a linear function such that for all x, f(3x+3)=4x+1. Find F(x)

I dont even know where to begin...

I am confused...you use F(x) and f(3x + 3)....is there a difference between the function defined as F and the one defined as f?

Oh well....
You might start here. If F(x) is a linear function, then

F(x) = mx + b

and you know that F(3x + 3) = 4x + 1

So, 4x + 1 = m(3x + 3) + b

could that be of some help?

 

[rofl]

no they are the same im sorry...f(x) and f(3x+3) i didnt mean to capitalize that f. i got that far, but there are 3 variables so i cant solve for anything.

 

[skeeter]

there are only two values to find, m & b. ... and yes you can find them both.

4x + 1 = m(3x+3) + b

4x + 1 = 3mx + 3m + b

using the method of equating coefficients ...

3m = 4, and 3m + b = 1

now ... find m and b.

 

[stapel]

I dont even know where to begin...

...i got that far, but there are 3 variables so i cant solve for anything.

In the future, it might be helpful if, instead of saying that you can't even get started, you posted how far you had gotten. That way, the tutors can get you going from from wherever you're stuck, rather than (accidentally) wasting your time by repeating what (you later say) you've already done.

Thank you for your consideration.

Eliz.

 

[soroban]

Hello, rofl!

Let \(\displaystyle f(x)\) be a linear function such that for all \(\displaystyle x,\:f(3x+3)\:=\:4x\,+\,1.\;\) Find \(\displaystyle F(x)\)

Since \(\displaystyle f(x)\) is a linear function, let \(\displaystyle f(x)\:=\:ax\,+\,b\)

Since \(\displaystyle f(3x+3)\:=\:4x\,+\,1\), we have: \(\displaystyle \:3(ax\,+\,b)\,+\,3\:=\:4x\,+\,1\)

. . Hence: \(\displaystyle \:3ax\,+\,(3b\,+\,3)\;=\;4x\,+\,1\)


Two polynomials are equal if their corresponding coefficients are equal.
. . \(\displaystyle \begin{array}3a\:=\:4 &\;\; \Rightarrow \;\;& a\,=\,\frac{4}{3} \\ 3b\,+\,3\:=\:1 &\;\; \Rightarrow\;\; & b\:=\:-\frac{2}{3}\end{array}\)


Therefore: \(\displaystyle \:f(x)\:=\:\frac{4}{3}x\,-\,\frac{2}{3}\)