Calculus question about the existence of antiderivativeLet f(x) be the function equal 0 on R &lt; 0 and equal 1 otherwise. Is clear to me that f has derivative undefined at x = 0. Is intuitive because the derivative would be ∞ as f increases by 1 in an infinitely small space.Now I was wondering whether antiderivative exists at 0. My thoughts: antiderivative corresponds to area under the function. Before zero the are is always zero and after zero the area is 1 so that like in derivative case the area increases by 1 in an infinitely small space. But I don't know if intuition can be used in this case.When does antiderivatives of functions exists in general and when not?

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Calculus question about the existence of antiderivativeLet f(x) be the function equal 0 on             R              &amp;lt;      0       and equal 1 otherwise. Is clear to me that f has derivative undefined at   x  =  0. Is intuitive because the derivative would be   ∞ as f increases by 1 in an infinitely small space.Now I was wondering whether antiderivative exists at 0. My thoughts: antiderivative corresponds to area under the function. Before zero the are is always zero and after zero the area is 1 so that like in derivative case the area increases by 1 in an infinitely small space. But I don&#039;t know if intuition can be used in this case.When does antiderivatives of functions exists in general and when not?

Schetterai · Accepted Answer

Step 1The two correct answers so far invoke the intermediate value property of derivatives, so I thought it might be useful to give an alternative argument avoiding that fact. Notice first that your function f is (Riemann) integrable over any finite interval. In particular, its integral from -1 to x is 0 for   x  ≤  0 and x for   x  ≥  0.Step 2If f had an antiderivative F, then, by the fundamental theorem of calculus, F would have to agree with this integral, up to an additive constant of integration. That is, there would be a constant C such that F(x) equals C for   x  ≤  0 and equals   x  +  C for   x  ≥  0. But then the derivative of F at zero doesn&#039;t exist, so F can&#039;t be the antiderivative of anything.

Finley Mckinney · Accepted Answer

Step 1To be sure, the assertion that for any differentiable function   f  :  I  →      R  , the derivative       f    ′    :  I  →      R   has the intermediate value property (whether it is continuous or not!) is a theorem of J.G. Darboux. For a careful statement and proof, see e.g.Step 2The basic idea is simple: one can reduce to the case in which   a  &amp;lt;  b  ∈  I,       f    ′    (  a  )  &amp;lt;  0,       f    ′    (  b  )  &amp;gt;  0, and one wants to show that there is a c with   a  &amp;lt;  c  &amp;lt;  b such that       f    ′    (  c  )  =  0. In fact the conditions       f    ′    (  a  )  &amp;lt;  0 and       f    ′    (  b  )  &amp;gt;  0 imply that the function f cannot have a minimum value at either a or b. Since f is differentiable, f is continuous and takes a minimum somewhere on [a,b], hence necesssarily at an interior point c, and then we must have       f    ′    (  c  )  =  0.

Calculus question about the existence of antiderivative Let f(x) be the function equal 0 on

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