Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am having trouble with this:

If $h(x)=0$ for $x<0$ and $h(x)=1$ for $x\geq 0$, prove there does not exist a function $f:\mathbb{R}\to \mathbb{R}$ such that $f'(x)=h(x)$ for all $x \in \mathbb{R}$. Give examples of two functions, not differing by a constant, whose derivatives equal $h(x)$ for all $x\neq 0$.

For the last part, Obviously if $h(x) = 1$ for $x<0$ and $h(x)=x$ for $x>0$ it will work. Is there another example? Also, I am not sure where to begin on the proof.

share|cite|improve this question
What about the function f:R->R such that f(x)=0 for all x<=0 and f(x)= x for all x>0 – Amr Nov 9 '12 at 0:44
To show that such a function doesn't exist, you could look at the difference quotient for the derivative at zero: $\lim_{h \to 0^-} \frac{f(h)-f(0)}{h}$. Since the derivative is zero for $x < 0$, $f(h)$ is constant for all $h< 0$. By continuity, you must have $f(h)= f(0)$... – ec92 Nov 9 '12 at 1:45
up vote 4 down vote accepted

All derivatives satisfy the intermediate value property. If they assume two values on an interval, the assume all of the values in between. This is a consequence of the Mean Value Theorem. Hence, your function is doomed.

share|cite|improve this answer
This is a good way to look at the problem! No neede to do anything, since the given $f'(x)$ doesn't satisfy IVT. +1. – coffeemath Nov 9 '12 at 2:18
I think this theorem is also known as Darboux's theorem. In the proof the mean value theorem is applied to something like $f(x)-x$. – nayrb Nov 9 '12 at 2:27
All derivatives defined on an interval. As coffeemath shows, there are many fine $f$'s that satisfy the requirement except for not having a proper derivative at zero. – Ross Millikan Nov 9 '12 at 2:42
By the way he phrases the question, he is looking for a function whose derivative exists throughout. – ncmathsadist Nov 9 '12 at 2:43

For the last part, you can define $f(x)=k_1$ for $x<0$ and $f(x)=x+k_2$ for $x>0$, and let $f(0)$ be any constant. Since the $k_1,k_2$ are arbitrary, there are infinitely many choices.

Note that here with different choices say $(1,2)$ and $(3,5)$ for $(k_1,k_2)$ you'll have functions not differing by a constant, in the sense that the differences will be different constants for $x<0$ than for $x>0$.

For the first part, suppose $f(-1)=c$ and use that $$f(x)=f(-1)+\int_{-1}^x f'(t)dt.$$ Then on doing the integral you find that $f(x)=c$ for x<0 and $f(x)=c+x$ for $x \ge 0$. This function is continuous, but has a "corner" at $x=0$ so is not differentiable on the whole set of reals.

share|cite|improve this answer
If anyone read this already, note that I forgot the prime on the derivative $f'(t)$, which I've just inserted. – coffeemath Nov 9 '12 at 2:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.