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This problem deals with strict monotonicity.

Example that $f$ can be strictly decreasing on an interval and yet have a derivative of zero at one or more places?

What about the converse? If $f' < 0$ on $(a,b)$, must $f$ be strictly decreasing on $(a,b)$?

big_math_boy
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1 Answers1

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$f(x) = -x^3$ on $(-1,1)$.

If $f'(x)<0$ on $(a,b)$, then yes, it must be strictly decreasing. Play around with the limit definition of the derivative to find out why.

user555203
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  • Thank you for the example. For the limit definition of the derivative, i'm thinking i use $\lim_{y \to x} \frac{f(y)-f(x)}{y-x}$ and if I think about the definition of strictly decreasing ($x<y$ implies $f(x)>f(y)$), then we know the fraction will be a negative value over a positive value, so it will turn out that the limit definition of the derivative is less than zero. Is that what you also had in mind? – big_math_boy Feb 05 '19 at 01:36
  • Pretty much, although you have to go the other way around. You are assuming $f'(x)<0$ and need to show $f$ is decreasing. If there's a sequence of points $(x_n)$ converging to $x$ from the right so that $f(x)\le f(x_n)$ for all $n$, then the limit for the derivative cannot be negative. You can do a similar thing when approaching from the left. That will tell you that there's a small interval $(x-\delta,x+\delta)$ on which $f$ is decreasing. And that holds for all $x$. Good luck! – user555203 Feb 05 '19 at 01:50
  • The result you mention is based on mean value theorem and not just definition of derivative. Also if $f'(x) <0$ it is not necessary that $f$ is decreasing in some interval of the form $(x-\delta, x+\delta) $. Classic counter example is $f(x) =x^2\sin(1/x)-x/2,f(0)=0$ so that $f'(0)=-1/2$ but $f$ is not monotone on any interval containing $0$. – Paramanand Singh Feb 05 '19 at 04:58
  • You're right -- that's fair. What I wanted to say was that there's a small interval $(x-\delta,x+\delta)$ so that $f(y)<f(x)$ for $x-\delta<y<x$ and $f(x)<f(y)$ for $x<y<x+\delta$. I incorrectly tried to say that concisely.

    However, I wouldn't say the result I'm stating is based on the mean value theorem. The argument I suggest is essentially how the MVT is proved, anyway. You can prove this result with very little more work than I suggested.

     – user555203 Feb 05 '19 at 13:49