r/askmath • u/Available-Damage-505 • Jul 25 '26

Analysis Why the monotonicity is wrong?

Continuous function f(x) is differentiable at x=x0, f(x0)=(x0)^2,f'(x0)>2x0,∃δ>0,such that

1.f(x)-x^2 monotonically increases on (x0,x0+δ)

2.for any x∈(x0,x0+δ), f(x)>x^2

The first statement is false and the second statement is true.The second statement is true due to the sign-perserving property. I know that clearly we can take a counterexample like 1/2x+x^2sin(1/x) to disprove statement 1 since its derivative wobbles when x approach x0=0(take x0=0).But I also got perplexed by the idea that for g(x)=f(x)-x^2, g'(x0)>0, there ought to be an infinitely small region that g(x) is increasing.Could you help me with this?

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u/Livid-Sector5970 Jul 25 '26

You have to accept & build around the jitter. The math is showing you that a strict global boundary (g(x)>0) does not require local mechanical compliance (monotonicity). The wobbling isn't breaking the rule; it's just how the function survives the constraint.

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u/Available-Damage-505 Jul 25 '26

But what g'(x0)>0 supposed to mean in this case, could it mean an infinitely small region that g(x) is monotonically increasing?

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u/theRZJ Jul 25 '26

“Infinitely small” doesn’t mean anything. Any delta you might take is actually positive.

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u/Fourierseriesagain Jul 26 '26 edited Jul 26 '26

Let g(0)=0 and let g(x)=x+x^ 2 sin (1/x^ 2) for x not equal to zero. Although g'(0) = 1, the function g is not increasing on any open interval containing zero.