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helppppp

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Let $$f(x) = \begin{cases} k(x) &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}$$Find the function k(x) such that f  is its own inverse.

What do they mean by "k(x) such that f  is its own inverse."

Aug 31, 2018

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Look at Melody's graph here, Mathtoo : https://www.desmos.com/calculator/fpruev1hep

Note that that all the coordinates  (a, b)  on the graph have a corresponding point (b, a) on the  graph

For instance....note that the point  (0,6)  is on the graph....and note also that  the point  (6, 0)  is also on the graph

[ The inverse  of a function just reverses the coordinates produced by that function ]

Does that make sense  ???

Aug 31, 2018
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Ok, and k(x) has to be greater than  2?

Aug 31, 2018
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Correct.....!!!

CPhill  Aug 31, 2018
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Thanks!

mathtoo  Aug 31, 2018
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Another way to understand this question is this:

an "inverse" of a function f is a function f-1 such that for every number x, f(f-1(x))=x=f-1(f(x)). Some functions have an inverse functions and others don't.

So the meaning of "f is its own inverse" is that for every x, f(f(x))=x.

Aug 31, 2018
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And, what do you do if you know k(x) is greater than 2.

Aug 31, 2018
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k(x) is not greater than 2, it's a function that is defined for x>2 (k(x) is defined only when x is greater than 2)

Guest Aug 31, 2018
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Do you plug a value of x? Like, 6?

Aug 31, 2018
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try to substitute a number smaller than 2 for x in the equation f(f(x))=x and see what happens.

example: when we substitute 1 for x we get:

f(f(1))=f(3) (we know that f(1)=3 because we defined the function f for 2 and values that are smaller than 2 by f(x)=2+(x-2)2)

so:

f(f(1))=f(3)=1 (we know that f(f(1))=1 because f is it's own inverse, meaning that f(f(x))=x)

but we know that for values of x that are larger than 2, f(x)=k(x). 3 is larger than 2, so now we know that 1=f(3)=k(3)

so k(3)=1. Try to use that way to find k(x) for every value of x that is larger than 2.

nvm i see you got it

Guest Aug 31, 2018
edited by Guest  Aug 31, 2018
edited by Guest  Aug 31, 2018
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Oh, I got it! Thanks!

Aug 31, 2018