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1. Let $f$ be defined by\(f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\)Calculate \(f^{-1}(0)+f^{-1}(6).\)

2. There are numbers A and B for which\(\frac A{x-1}+\frac B{x+1}=\frac{x+2}{x^2-1}\)for every number \(x\neq\pm1\). Find A-B.

 

3.The greatest integer function, \(\lfloor x\rfloor\), denotes the largest integer less than or equal to x. For example, \(\lfloor3.5\rfloor=3\), \(\lfloor\pi\rfloor=3\) and \(\lfloor -\pi\rfloor=-4\). Find the sum of the three smallest positive solutions to \(x-\lfloor x\rfloor=\frac1{\lfloor x\rfloor}.\) Express your answer as a mixed number.

 

thank you in advance

 Sep 16, 2019
 #1
avatar+111401 
+2

2)    x + 2                 A           B

      ______  =       ____ +   ____        

       x^2 - 1            x - 1       x + 1

 

Factor the denominator of the first fraction as  ( x -1) (x + 1)   and multiply through by this factorization and we get that

 

x + 2  = A ( x + 1)   +  B ( x - 1)      simplify

 

1x + 2  =  (A + B)x  + ( A - B)        equate terms and we have that

 

A +  B  =  1

 

A -  B   =  2         

 

 

cool cool cool

 Sep 16, 2019
 #2
avatar+111401 
+2

1) 

 

f-1 (0)    means that when we plug  an original x value into one of the functions, then it returns  0

 

Note  that   in the first function if we put 3 into it for x, we get  3 - x =   3   - 3  = 0

And  x = 3  is in the domain of this function ...... so 

 

f-1(0)  = 3

 

 

f-1(6)    means that when we plug an original x value   into one of the functions, then it returns 6

 

Note that if we put  -3  into the first function for x, we get that  3 - x   = 3 - (-3)  = 6

And x = -3  is in the domain of this function, as well....so

 

f-1(6)  =-3

 

So

 

f-1(0)  + f-1(6)   = 3   + (-3)     =    0

 

 

cool cool cool

 Sep 16, 2019
 #3
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0

THank you

 Sep 16, 2019

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