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May 22nd, 2014, 10:38 AM   #1
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Square Roots

Can someone help me out here. What is the difference between -√(x-2)^2 and √(x-2)^2. I'm using khan academy to learn algebra 2 and this showed up on one of the problems to make a function inverse. Link to the problem itself below.

https://www.khanacademy.org/math/alg...rses-example-3
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May 22nd, 2014, 01:26 PM   #2
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Well there's a big difference between the two expressions.

If x=3,

$\displaystyle -\sqrt { { \left( 3-2 \right) }^{ 2 } } =-\sqrt { 1 } =-1$

$\displaystyle \sqrt { { \left( 3-2 \right) }^{ 2 } } =\sqrt { 1 } =1$

Take a look at this graph: https://www.desmos.com/calculator/qohzruuclo
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May 22nd, 2014, 02:04 PM   #3
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Ok...

$\displaystyle f\left( x \right) =y={ \left( x-1 \right) }^{ 2 }-2\quad \{ x\le 1\} \\ \\ When\quad x=1,\quad y=-2\\ \\ When\quad x\rightarrow \infty ,\quad y\rightarrow \infty \quad \therefore \quad f\left( x \right) \ge -2\\ \\ Now...\\ \\ y={ \left( x-1 \right) }^{ 2 }-2\\ \\ { \left( x-1 \right) }^{ 2 }=y+2\\ \\ x-1=\pm \sqrt { y+2 } \\ \\ x=\pm \sqrt { y+2 } +1\\ \\ \therefore \quad f^{ -1 }\left( x \right) =\sqrt { x+2 } +1\quad or...\quad f^{ -1 }\left( x \right) =-\sqrt { x+2 } +1\\ \\ but\quad domain\quad of\quad f^{ -1 }\left( x \right) \quad is\quad \{ x\ge -2\} ,\quad and\quad f^{ -1 }\left( x \right) \le 1\\ \\ \sqrt { x+2 } +1\le 1\\ \\ \sqrt { x+2 } \le 0\\ \\ x+2\le 0\\ \\ x\le -2\quad but\quad domain\quad of\quad f^{ -1 }\left( x \right) \quad is\quad \{ x\ge -2\} \\ \\ \therefore \quad f^{ -1 }\left( x \right) =-\sqrt { x+2 } +1\quad \{ x\ge -2\} \quad f^{ -1 }\left( x \right) \le 1\\ \\$

Graph Demonstration: https://www.desmos.com/calculator/1vh56wxriv
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