1. ## Function math problem.

Hello, again!

Is given: $f\left(\frac{2x-1}{x+1}\right)=x^2-3x+4\;\;\;\text{find:}\;\;f(1)$

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I've done tasks like this one before, like: $f(x-2)=x^3-2x-1\;\;\;,\;\;\text{find:}\;f(1)$

$A)\;11$
$B)\;9$
$C)\;20$
$D)\;31,5$

$t=x-2\;\;=>\;x=t+2$

$f(t)=(t+2)^3-2(t+2)-1$

$f(t)=t^3+3t^2*2+3t2^2+2^3-2t-4-1$

$f(t)=t^3+6t^2+10t+3$

$f(x)=x^3+6x^2+10x+3$

$f(1)=1^3+6*1^2+10*1+3$

$f(1)=7+13$

$f(1)=20$

The attempt at a solution:

$f\left(\frac{2x-1}{x+1}\right)=x^2-3x+4\;\;\;\text{find:}\;\;f(1)$

$t=\frac{2x-1}{x+1}$

$t(x+1)=2x-1$

$tx+t=2x-1$

$tx+t+1=2x$

As you can see, I'm stuck in the beginning...how to separate X in one side??
Thank you!

2. ## Re: Function math problem.

By inspection, you need to substitute $x=2$ to find $f(1)$.

3. ## Re: Function math problem.

Originally Posted by kreshnik
Hello, again!

Is given: $f\left(\frac{2x-1}{x+1}\right)=x^2-3x+4\;\;\;\text{find:}\;\;f(1)$
Solve: $\frac{2x-1}{x+1}=1$

4. ## Re: Function math problem.

Originally Posted by Siron
Solve: $\frac{2x-1}{x+1}=1$
I guess substituting t with 1 is easier.

$\frac{2x-1}{x+1}=1\;\;\rightarrow\;\;x=2$

$f(1)=1^2-3*1+4$

$f(1)=1+1$

$f(1)=2$

$A)\;0\;\;\;\;B)\;1\;\;\;\;C)\;2\;\;\;\;D)\;3$

Thank you very much everyone.