Define a function f such that for x ≠ 0 f ( x ) = c o s ( 1 / x ) and f ( 0 ) = 1 / 2. does it admits antiderivatives?

Question

Define a function f such that for   x  ≠  0  f  (  x  )  =  c  o  s  (  1      /    x  ) and   f  (  0  )  =  1      /    2. does it admits antiderivatives?

Tianna Deleon · Accepted Answer

Step 1Hint: See if you can prove this: if f is bounded, and for every a with   0  &amp;lt;  a  &amp;lt;  1, f is Riemann integrable on [a,1], then f is Riemann integrable on [0,1]. Do the same thing for   [  −  1  ,  b  ],   −  1  &amp;lt;  b  &amp;lt;  0.If there is any anti-derivative of f, then this would be it:  F  (  x  )  =      ∫    0    x    f  (  t  )    d  t  ,    for all real numbers   xStep 2If f is continuous except at 0, this gives us F with       F    ′    (  x  )  =  f  (  x  ) for all x except possibly   x  =  0.In the particular case on the OP, we get       F    ′    (  0  )  =  0, but I will not prove that here. So your answer is &quot;no&quot;, your function has no anti-derivative on [-1, 1] .The Darboux property is necessary, but not sufficient, for the existence of an anti-derivative.

Define a function f such that for x ≠ 0 f ( x ) = c o s

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