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Consider the function $f(x) = x^3$. We have $f'(x) = 3x^2$. So, $f'(x) > 0$ for all $x \ne 0$. So, $f(x)$ is increasing on any open interval $I$ which doesn't include zero. Now, for some $\varepsilon > 0$, consider the open interval $(-\varepsilon, \varepsilon)$. Can it be said that $f$ is increasing on this interval even though $f'(0) = 0$ ?

user9026
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    It's even strictly increasing, in the sense that $a>b\implies f(a)>f(b)$. – lulu Oct 07 '24 at 10:25
  • I understand that. But why $f(x)$ is increasing on $(-\varepsilon, \varepsilon)$ even though there is a point in this interval where $f'(x) = 0$ ? I am using Stewart's Calculus, so may be definitions there are not rigorous enough. – user9026 Oct 07 '24 at 10:29
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    You should use your terms carefully, or define them. An Increasing Function is one that satisfies $a>b\implies f(a)≥f(b)$. A strictly increasing function, like $x^3$, satisfies the stronger relation $a>b\implies f(a)>f(b)$. Nothing to do with derivatives. It is true, of course, that if we have a differentiable function for which $f'(x)>0$ for all $x$, then that function is strictly increasing, but the converse is not true. – lulu Oct 07 '24 at 10:32

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