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Consider the function $f$ defined on the nonnegative reals such that $f(x)=x$ for all nonnegative $x$ and the function $g$ defined on the nonnegative reals such that $g(0)=0$, $g^{\prime}(0)=987654321$, and $g^{\prime}(x) = 1$ for all $x \in \mathbb{R}^+$.

Are these functions equivalent? Why or why not?

yuri12345
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    What is your notion of "equivalent"? – Eric Towers Oct 26 '19 at 21:35
  • I suppose I am wondering whether $f(x)-g(x)=0$ for all nonnegative $x$. – yuri12345 Oct 26 '19 at 21:36
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    How can $f$ and $g$ be equal if $g'(0) = 987654321$ and $f'(0) = 1$? – Klaus Oct 26 '19 at 21:40
  • I think if you define a function only on the nonnegative reals then that means a derivative in zero can't exist, because per definition you'd need to be able to take a limit on both ends of 0. So if I'm correct your definition of $g$ makes no sense. – WafflesTasty Oct 26 '19 at 21:40
  • They are not equivalent. The function $g(x)$ must have some other definition at $x=0$ so that after taking the first derivative it has the assigned value. – Dinno Koluh Oct 26 '19 at 21:40
  • What if instead $g^{\prime}(1)=987654321$ and $g^{\prime}(x)=1$ for all nonnegative $x \neq 1$? Then would $f(x)=g(x)$ for all nonnegative $x$? – yuri12345 Oct 26 '19 at 21:52

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Such a function $g$ does not exist, since $g'$ cannot have a jump discontinuity. See the question below, or Darboux's theorem.

Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.

Milten
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