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Given that f(2) = 5  and \({f}^{-1}(x+4) = 2{f}^{-1}(x) + 1\) for all x, find \({f}^{-1}(17)\)

 Mar 19, 2017
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f-1​(13 + 4) = 2f-1​(13) + 1

f-1​(17) = 2f-1​(13) + 1

 

f-1​(9 + 4) = 2f-1​(9) + 1

f-1​(13) = 2f-1​(9) + 1

 

f-1​(5 + 4) = 2f-1​(5) + 1

f-1​(9) = 2f-1​(5) + 1

 

f(2) = 5, so f-1(5) = 2

Substitute:

f-1​(9) = 2(2) + 1 = 5

 

f-1​(13) = 2(5) + 1 = 11

 

f-1​(17) = 2(11) + 1 = 23

 Mar 20, 2017

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