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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 $f^{-1}(0)+f^{-1}(6)$.

Guest Sep 23, 2017
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\( \[f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\]Calculate $f^{-1}(0)+f^{-1}(6)$.\)

 

f-1 (0)    must mean   that either   3 - x  = 0       or   -x^3 + 2x^2  + 3x  = 0

However......in the second function x = 0    is the only thing that makes this true, but that function isn't defined for x  = 0

So.....it must mean that  3 - x  = 0....and x  = 3......so   f-1 (0)   = 3

 

And

 

f-1 (6)  must mean that either   3 - x  = 6      or that    -x^3 + 2x^2  + 3x  = 6

Looking at the second function  we have that

-x^3 +2x^2  + 3x  = 6  →    x^3 -2x^2 - 3x + 6  =  0  →   x^2( x - 2) - 3(x - 2)  = 0  →

(x^2 - 3) (x -2)  = 0  →    x  = ±√3   or x  = 2   are the x values that make this true....but....these values are all < 3  and this function isn't defined for these values

But    3 - x   = 6     when  x  = -3     which is valid

 

So

 

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

 

So  ....  their sum   = 0 

 

 

 

cool cool cool

CPhill  Sep 23, 2017

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