Find the inverse of a piecewise function, namely: f ( x ) = { x − 1 , 0 ≤ x &lt; 1 2 − x , 1 &lt; x ≤ 2 For the first case f : y = x − 1 when 0 ≤ x &lt; 1 → f − 1 : x = y + 1 when − 1 ≤ x &lt; 0But I find this because I only have to subtract 1 from the endpoints of the inequality. I don't have a clear intuition how to do this with 2−x

Question

Find the inverse of a piecewise function, namely:  f  (  x  )  =      {                            x          −          1          ,                          0          ≤          x          &amp;lt;          1                                      2          −          x          ,                          1          &amp;lt;          x          ≤          2                        For the first case   f  :  y  =  x  −  1 when   0  ≤  x  &amp;lt;  1  →      f          −      1        :  x  =  y  +  1 when   −  1  ≤  x  &amp;lt;  0But I find this because I only have to subtract 1 from the endpoints of the inequality. I don&#039;t have a clear intuition how to do this with 2−x

lascosasdeali3v · Accepted Answer

Suppose that       f          −      1       there exists, then you have   f  (  x  )  =  x  −  1 when   0  ⩽  x  &amp;lt;  1 then setting   y  =  f  (  x  ) we get   y  =  x  −  1 when   0  ⩽  x  &amp;lt;  1 then   x  =  y  +  1 when   0  ⩽  y  +  1  &amp;lt;  1. Hence       f          −      1        (  x  )  =  x  +  1 when   −  1  ⩽  x  &amp;lt;  0.Also   f  (  x  )  =  2  −  x when   0  &amp;lt;  x  ⩽  2 then setting y=f(x) we get   y  =  2  −  x when   1  &amp;lt;  x  ⩽  2 then   x  =  2  −  y when   1  &amp;lt;  2  −  y  ⩽  2. Hence       f          −      1        (  x  )  =  2  −  x when   0  ⩽  x  &amp;lt;  1Therefore, the inverse of function f is given by                                          f                          −              1                                :          x          ↦                      {                                                            x                  +                  1                  ,                                    if                                    −                  1                  ⩽                  x                  &amp;lt;                  0                  ,                                                                              2                  −                  x                  ,                                    if                                    0                  ⩽                  x                  &amp;lt;                  1                                                                                    Now check   (      f          −      1        ∘  f  )  (  x  )  =  x for all   x  ∈      D    o    m    (      f          −      1        ) and   (  f  ∘      f          −      1        )  (  x  )  =  x for all   x  ∈      D    o    m    (  f  )

Find the inverse of a piecewise function: f(x)={(x-1,0<=x<1),(2-x,1<x<=2):}

Answered question

Answer & Explanation

New Questions in Algebra I