Find f^-1(x) <-- inverse if f(x) = x^2 - 4x + 3, x <= 2 find value of f^-1(8).

I got x=29 BUT x<= 2????

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- Oct 30th 2008, 12:37 PMmwokInverse of function WITH a limit??
Find f^-1(x) <-- inverse if f(x) = x^2 - 4x + 3, x <= 2 find value of f^-1(8).

I got x=29 BUT x<= 2???? - Oct 30th 2008, 12:53 PMmasters
- Oct 30th 2008, 01:00 PMmwok
- Oct 30th 2008, 01:12 PMmasters
Not quite. Your original function was $\displaystyle f(x)=x^2-4x+3$

$\displaystyle f(x)=y=x^2-4x+3$

Interchange**x**and**y**

$\displaystyle x=y^2-4y+3$

Isolate the**y**

$\displaystyle y^2-4y=x-3$

Complete the square on the left side. You'll see why in a minute.

$\displaystyle y^2-4y+4=x-3+4$

$\displaystyle (y-2)^2=x+1$

$\displaystyle y-2=\pm\sqrt{x+1}$

$\displaystyle y=\pm\sqrt{x+1}+2$

$\displaystyle \boxed{f^{-1}(x)=\pm\sqrt{x+1}+2}$ - Oct 30th 2008, 01:18 PMmwok
Wow! I didn't know you could plus 4 to each side and solve the problem like that. Are there multiple ways to solve this?

- Oct 30th 2008, 01:28 PMmasters
- Oct 30th 2008, 01:30 PMmwok