This means that if we know a point on one side of the

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Unformatted text preview: se should be, © 2007 Paul Dawkins 201 http://tutorial.math.lamar.edu/terms.aspx College Algebra h -1 ( x ) = 4 + 5x 2x -1 Finally we’ll need to do the verification. This is also a fairly messy process and it doesn’t really matter which one we work with. ( h o h ) ( x ) = h é h ( x )ù ë û -1 -1 é 4 + 5x ù = hê ë 2x -1 ú û 4 + 5x +4 2x -1 = æ 4 + 5x ö 2ç ÷-5 è 2x -1 ø Okay, this is a mess. Let’s simplify things up a little bit by multiplying the numerator and denominator by 2 x - 1 . 4 + 5x +4 2x -1 2x -1 h o h -1 ) ( x ) = ( 2 x -1 æ 4 + 5x ö 2ç ÷-5 è 2x -1 ø 4 + 5x ö + 4÷ ( 2 x - 1) æ ç è 2x -1 ø = æ æ 4 + 5x ö ö ( 2 x - 1) ç 2 ç ÷ - 5÷ è è 2x -1 ø ø 4 + 5 x + 4 ( 2 x - 1) = 2 ( 4 + 5 x ) - 5 ( 2 x - 1) 4 + 5x + 8x - 4 8 + 10 x - 10 x + 5 13 x = 13 =x = Wow. That was a lot of work, but it all worked out in the end. We did all of our work correctly and we do in fact have the inverse. There is one final topic that we need to address quickly before we lea...
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