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(a) Must there be an interval around $0$ on which the derivative remains positive?

(b) Is there an open interval around $0$ on which $f$ is increasing?

For (a) I believe the answer is yes because $f$ is differentiable and the derivative $f'(0)$ exists. Since $f'(x) = lim_{x \rightarrow 0} \dfrac{f(x)-0}{x-0}>0$, then

$lim_{x \rightarrow 0-} \dfrac{f(x)-0}{x-0}>0$ and $lim_{x \rightarrow 0+} \dfrac{f(x)-0}{x-0}>0$

Am I on the right track here?

As far as (b) is concerned, I thought I would use the Mean Value Theorem and define the interval around $c>0$, but I know I can't start with point $c$ and from there find an interval $(a,b)$, so maybe there's another way, like Darboux's Theorem?

Thanks in advance!

Jabernet
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  • For (a), take the function in http://math.stackexchange.com/questions/292275/discontinuous-derivative and add $x$. This gives a counterexample. – Henricus V. Feb 10 '17 at 05:20
  • No to both questions. Answers will be yes if $f'$ is continuous at $0$. Hence one must construct a counter-example example where $f'$ is discontinuous at $0$. – Paramanand Singh Feb 10 '17 at 05:43

2 Answers2

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The counter-example from user detnvvp's answer shows that the answer to both your queries is NO. I want to address some concerns over your approach in this answer.

You start off with $$f'(x) = \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0} > 0$$ I think there is a typo and the correct equation is $$f'(0) = \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0} > 0$$ And then you show that the corresponding one sided limits are also positive. From that you get two things (which you should have tried to work out):

1) There is an $h > 0$ such that $f(x) > f(0)$ for all $x \in (0, h)$.

2) There is a $k > 0$ such that $f(x) < f(0)$ for all $x \in (-k, 0)$.

Such a behavior is expressed by saying that $f$ is strictly increasing at point $0$. But note that this is not the same as saying that $f$ is increasing in some interval containing $0$. Precisely we can't combine 1), 2) above to say that $f$ is strictly increasing in $(-k, h)$.

Thus you should not expect the answer of $(b)$ to be affirmative.

Regarding $(a)$ note that continuous functions have sign preserving property i.e. if $f$ is continuous at $a$ and $f(a) \neq 0$ then $f$ has same sign as that of $f(a)$ in some interval containing $a$.

So the answer to $(a)$ is affirmative if $f'$ is continuous at $0$. But we are not given this so we should not expect an affirmative answer for $(a)$.

Paramanand Singh
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(b) is wrong: to see that, consider the function $$f(x)=\left\{\begin{array}{c l}x^2\sin\frac{1}{x}+\frac{1}{2}x, & x\neq 0 \\ 0, & x=0\end{array}\right..$$ Then $f'(0)=\frac{1}{2}>0$, but $f$ is not increasing on a neighborhood of $0$. To see that, consider the derivative at any $x>0$: $$f'(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}+\frac{1}{2},$$ which changes sign infinitely often as $x\to 0^+$.

detnvvp
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