# Function math problem.

• Jan 15th 2012, 12:02 PM
kreshnik
Function math problem.
Hello, again!

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

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

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

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

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

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

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

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

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

$\displaystyle f(1)=7+13$

$\displaystyle f(1)=20$

The attempt at a solution:

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

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

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

$\displaystyle tx+t=2x-1$

$\displaystyle tx+t+1=2x$

As you can see, I'm stuck in the beginning...how to separate X in one side??
Thank you!
• Jan 15th 2012, 12:26 PM
alexmahone
Re: Function math problem.
By inspection, you need to substitute $\displaystyle x=2$ to find $\displaystyle f(1)$.
• Jan 15th 2012, 12:33 PM
Siron
Re: Function math problem.
Quote:

Originally Posted by kreshnik
Hello, again!

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

Solve: $\displaystyle \frac{2x-1}{x+1}=1$
• Jan 15th 2012, 11:42 PM
kreshnik
Re: Function math problem.
Quote:

Originally Posted by Siron
Solve: $\displaystyle \frac{2x-1}{x+1}=1$

I guess substituting t with 1 is easier.

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

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

$\displaystyle f(1)=1+1$

$\displaystyle f(1)=2$

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

Thank you very much everyone.(Hi)