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Unformatted text preview: x ) . f 1 ( x ) = x2
+
33 Now, we need to verify the results. We already took care of this in the previous section, however,
we really should follow the process so we’ll do that here. It doesn’t matter which of the two that
we check we just need to check one of them. This time we’ll check that f o f 1 ( x ) = x is ( true. ) ( f o f ) ( x ) = f é f ( x )ù
ë
û
1 1 é x 2ù
=fê + ú
ë3 3û
æ x 2ö
= 3ç + ÷  2
è 3 3ø
= x+22
=x
Example 2 Given g ( x ) = x  3 find g 1 ( x ) , x ³ 0 .
Solution
Now the fact that we’re now using g ( x ) instead of f ( x ) doesn’t change how the process
works. Here are the first few steps. y = x 3
x= y 3 Now, to solve for y we will need to first square both sides and then proceed as normal. x= y 3 x = y 3
2 x +3= y
2 This inverse is then, g 1 ( x ) = x 2 + 3 Finally let’s verify and this time we’ll use the other one just so we can say that we’ve gotten both
down somewhere in an example.
© 2007 Paul Dawkins 200 http://tutorial.math.lamar.edu/terms.aspx College Algebra (g 1 o g ) ( x ) = g 1 é g ( x ) ù
ë
û
= g 1
= ( ( x3 x 3 ) 2 ) +3 = x 3+ 3
=x
So, we di...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
 Spring '12
 MrVinh

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