SOLUTION: If y=f(x) is such that f(2b-1)=4b^2 + 2b -3 for any real value b, find an expression for f(x). The answer is f(x)= x^2 + 3x -1. I divided 4b^2 + 2b -3 by 2b-1 which gives quot

Algebra ->  Functions -> SOLUTION: If y=f(x) is such that f(2b-1)=4b^2 + 2b -3 for any real value b, find an expression for f(x). The answer is f(x)= x^2 + 3x -1. I divided 4b^2 + 2b -3 by 2b-1 which gives quot      Log On


   

Question 202011: If y=f(x) is such that f(2b-1)=4b^2 + 2b -3 for any real value b, find an expression for f(x).
The answer is f(x)= x^2 + 3x -1.
I divided 4b^2 + 2b -3 by 2b-1 which gives quotient of 2b + 2 remainder -1.
That's f(2b-1)= {(2b + 2)(2b -1)} -1 .
I don't know how to figure out (2b + 2). Please help. Thank you.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
f%282b-1%29=4b%5E2+%2B+2b+-3 Start with the given function


What we're seeing here is that when x=2b-1, then f%28x%29=4b%5E2+%2B+2b+-3.

To avoid any really technical jargon or complicated theory, what's basically happening is we're mapping f(x) to a new function g(b) when we plug in x=2b-1. Now we want to map back. How do we do that? We use the inverse of course. So basically, we need to solve for "b" in x=2b-1 to get b=%28x%2B1%29%2F2. Now plug it into the function g(b) to find f(x)

Note: if you plug in b=%28x%2B1%29%2F2 into 2b-1, we get 2%28%28x%2B1%29%2F2%29-1=x%2B1-1=x.




f%28x%29=4%28%28x%2B1%29%2F2%29%5E2+%2B+2%28%28x%2B1%29%2F2%29+-3 Plug in b=%28x%2B1%29%2F2


f%28x%29=4%28%28x%2B1%29%5E2%2F4%29+%2B+2%28%28x%2B1%29%2F2%29+-3 FOIL and square


f%28x%29=%284%28x%2B1%29%5E2%29%2F4+%2B+%282%28x%2B1%29%29%2F2+-3 Multiply


f%28x%29=%28x%2B1%29%5E2+%2B+x%2B1+-3 Reduce


f%28x%29=x%5E2%2B2x%2B1+%2B+x%2B1+-3 FOIL


f%28x%29=x%5E2%2B3x-1 Combine like terms.


So the original function is f%28x%29=x%5E2%2B3x-1


To verify that this is indeed the answer, simply plug in x=2b-1 and you should get f%282b-1%29=4b%5E2+%2B+2b+-3 again.