Finding a Function's Inverse

**shadow6** · April 6th 2010, 01:12 PM

$f(x) = 2(x - 1)^2 - 3$
$f(x) = (2x - 2)^2 - 3$
$(2x - 2)^2 = x + 3$
$2x - 2 = \pm \sqrt{x + 3}$
$2x = \pm \sqrt{\frac{x + 3}{2}}$
$x = \pm \sqrt{\frac{x + 3}{2}} + 2$
$f^-(x) = \pm \sqrt{\frac{x + 3}{2}} + 2$

That's how I solved it, but my textbook answer is
$f^-(x) = 1 \pm \sqrt{\frac{x + 3}{2}}$

Where was my error?

**skeeter** · April 6th 2010, 01:48 PM

Originally Posted by shadow6

$f(x) = 2(x - 1)^2 - 3$
$f(x) = (2x - 2)^2 - 3$ mistake in going from above line to here.

$\textcolor{red}{2(x-1)^2 \ne (2x-2)^2}$

Where was my error?

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