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#1**+1 **

Work backwards.

To get five on the last f(x), then x could be -3 for the first f(x).

Or, on the f(x)=x+3, it would be impossible.

So you need -3 after the first f(x).

To do that, you could start off with -6, get to -3 and get 5.

Or you could start off with x=1, get to -3, and get 5.

There are two possible values for x: \(\boxed{\{-6, 1\}}\) .

asdf1243 Jan 5, 2019

#2**+2 **

For f( f(x) to = 5

We have several possibilities

First

(x + 3) + 3 = 5

x + 6 = 5

x = -1

But.....note that f(f(-1)) = f (-3) = 5 so....-1 works

Also

(x^2 - 4)^2 - 4 = 5

x^4 - 8x^2 + 16 - 9 = 0

x^4 - 8x^2 + 7 = 0 factor

(x^2 - 7) ( x ^2 - 1)

We have 4 possible solutions

x = -√7, √7, -1 , 1

x = - √7 will produce because f(f(-√7)) = f (3) = 6

We have seen that x = - 1 works

x = 1 will work because f(f(1)) = f(-3) = 5

x = √7 will work because f(f(√7)) = f(3) = 5

Another possiibility is that

(x +3)^2 - 4 = 5

x^2 + 6x + 9 - 9 = 0

x^2 + 6x = 0

x(x + 6) = 0

x = 0 or x = -6

f(f(0)) = f(-4) = 12

f(f(-6)) = f(-3) = 5

The last possibility is that

(x^2 - 4) + 3 = 5

x^2 - 6 = 0

x = -√6 or x = √6

f(f(-√6)) = f (2) = 0

f(f(√6)) = f(2) = 0

So....the x values where f(f(x) ) = 5 are

x = -1, 1, -√7, √7, -6

CPhill Jan 6, 2019