Antiderivative of a branch functionLet f : R ↦ R , f ( x ) = e x for x ≤ 0 and a x + b for x &gt; 0. I want to determine the real numbers a and b for which f admits antiderivatives.I took a plausible antiderivative F ( x ) = e x + c for x &lt; 0, F ( 0 ) = α and a x 2 2 + b x + d for x &gt; 0 and using the fact that F must be first continuous and then differentiable, I showed that b = 1.This is easy to see as from continuity we must have that d = α = c + 1.Then, since F is clearly differentiable on R ∖ { 0 } if we impose the condition that the left and right derivatives at 0 of F to be equal so we must have: lim x → 0 , x &lt; 0 F ( x ) − F ( 0 ) x − 0 = lim x → 0 , x &gt; 0 F ( x ) − F ( 0 ) x − 0 Thus: lim x → 0 , x &lt; 0 e x + c − c − 1 x = lim x → 0 , x &gt; 0 a x 2 2 + b x + c + 1 − c − 1 x Which gives us b = 1.My question is then, does this suffice? Can a be chosen arbitrarily such that f has antiderivatives as long as b = 1. It would seem that the function F ( X ) = e x + c for x ≤ 0 and F ( x ) = a x 2 2 + x + c + 1 for x &gt; 0 for c ∈ R does indeed satisfy the necessary conditions.

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Antiderivative of a branch functionLet   f  :      R    ↦      R  ,   f  (  x  )  =      e          x       for   x  ≤  0 and   a  x  +  b for   x  &amp;gt;  0. I want to determine the real numbers a and b for which f admits antiderivatives.I took a plausible antiderivative   F  (  x  )  =      e          x        +  c for   x  &amp;lt;  0,   F  (  0  )  =  α and   a            x              2              2    +  b  x  +  d for   x  &amp;gt;  0 and using the fact that F must be first continuous and then differentiable, I showed that   b  =  1.This is easy to see as from continuity we must have that   d  =  α  =  c  +  1.Then, since F is clearly differentiable on       R    ∖  {  0  }  if we impose the condition that the left and right derivatives at 0 of F to be equal so we must have:      lim          x      →      0      ,      x      &amp;lt;      0                  F      (      x      )      −      F      (      0      )              x      −      0        =      lim          x      →      0      ,      x      &amp;gt;      0                  F      (      x      )      −      F      (      0      )              x      −      0      Thus:       lim          x      →      0      ,      x      &amp;lt;      0                          e                  x                    +      c      −      c      −      1        x    =      lim          x      →      0      ,      x      &amp;gt;      0                  a                        x                      2                          2            +      b      x      +      c      +      1      −      c      −      1        x  Which gives us   b  =  1.My question is then, does this suffice? Can a be chosen arbitrarily such that f has antiderivatives as long as   b  =  1. It would seem that the function   F  (  X  )  =      e          x        +  c for   x  ≤  0 and   F  (  x  )  =  a            x              2              2    +  x  +  c  +  1 for   x  &amp;gt;  0 for   c  ∈      R   does indeed satisfy the necessary conditions.

veceritzpzg · Accepted Answer

Explanation:By definition, f admits an antiderivative iff f is a derivative of some function. Thus, by Darboux theorem, f has to have an intermediate value property. For f of the given form this is equivalent to   b  =  1 as you got as well. Then, independently of the choice of a, f is continuous and has an antiderivative by (one of) the Fundamental Theorem(s) of Calculus.

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