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We have:

$$f(x)=\left\{\begin{matrix} -1, & -1 \leq x \leq 0 \\ 1, & 0<x \leq 1 \end{matrix}\right.$$

Why is the antiderivative equal to $F(x)=|x|-1$?

Do we have to find the integral of $f$ at both intervals? But... what limitsa of integration do we have to use??

EDIT: In my notes,there is the remark,that if:

$$f:[a,b] \to \mathbb{R} \text{ intergable }$$ $$F:[a,b] \to \mathbb{R}$$ $$F(x)=\int_a^x f$$ $$F'(x_0)=f(x_0), \text{ ONLY IF } x_0: \text{ point at which f is continuous }$$

As a counter-example,they give the function:

$f(x)=\left\{\begin{matrix} -1, & -1 \leq x \leq 0 \\ 1, & 0<x \leq 1 \end{matrix}\right.$ that is discontinuous at $0$,and then $F(x)=|x|-1$ is not differentiable at $0$.

evinda
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    The function $F(x)$ has the right derivative (except of course at $0$). There are others. I suspect you are not giving the full problem, which likely involves a definite integral. – André Nicolas Jun 23 '14 at 17:59
  • We cannot find an antiderivative $F$,that is differentiable at $0$,because $f$ is not continuous at this point,right?? – evinda Jun 23 '14 at 18:20
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    @evinda: Your argumentation is completely wrong. Of course a differentiable function is continuous, however that doesn't imply that a differentiable function has continuous derivative... – C-star-W-star Jun 23 '14 at 18:25
  • Could you explain it further to me? – evinda Jun 23 '14 at 18:28
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    Derivatives need not be continuous, so one cannot call the fact that $f$ is not continuous the reason. A fancy reason is that derivatives have the Intermediate Value property, which $f$ doesn't. A more basic reason is that the left and right limits of $f$ exist but are unequal. In a derivative, if the left and right limits exist, they must be equal. – André Nicolas Jun 23 '14 at 18:29
  • @AndréNicolas I understand...thank you very much!!! – evinda Jun 23 '14 at 23:44
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    You are welcome. – André Nicolas Jun 23 '14 at 23:47

1 Answers1

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Consider the function $F(x)=|x|-1$. If $-1\leq x<0$ then $F(x)=-x-1$. So the derivative of $F$ is $-1$ on $[-1,0)$. If $0<x\leq 1$ then $F(x)=x-1$. So the derivative of $F$ is $1$ on $(0,1]$.

We have shown $F'(x)=f(x)$ for all $x$ in the domain of $f$, i.e., $F$ is an antiderivative of $f$. Note that there may be others, such as simply $|x|$, or more generally $|x|+C$ for any $C\in \mathbb R$. As a consequence of the Mean Value Theorem, $|x|+C$ $C\in\mathbb R$ is the most general form that an antiderivative of $f$ can have.

Edit: If you include $0$ in the domain of $f$, then since $F$ is not differentiable at $0$, I would be inclined to say $F$ is not an antiderivative of $f$. However, if you only require that $F$ satisfy $F'(x)=f(x)$ where $f$ is continuous, $F$ still works, as $f$ is not continuous at $0$.

Forever Mozart
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  • Tom Cruise: Nice,I understand... So,we cannot find an antiderivative $F$ in a domain that includes also $0$,right?? – evinda Jun 23 '14 at 18:19
  • Correct. An antiderivative $F$ would have to be essentially $|x|$ on $[-1,0)\cup (0,1]$. To be differentiable at $0$ it would have to be continuous at $0$, meaning $F(0)$ completes the "V" shape. So $F$ would be $|x|$, which is not differentiable at $0$! – Forever Mozart Jun 23 '14 at 18:32
  • A ok...thanks!!! – evinda Jun 23 '14 at 23:45