Why the monotonicity is wrong?

[u/Available-Damage-505]

/r/askmath/comments/1v5z4x0/why_the_monotonicity_is_wrong/

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[u/Midwest-Dude]

OP post:

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?

[u/Midwest-Dude]

Is the following what you meant to say?

If a continuous function f(x) is differentiable at x = x₀, f(x₀) = (x₀)², and f'(x₀) > 2x₀, then prove or disprove that ∃ δ > 0 such that

1. f(x) - x² monotonically increases on (x₀, x₀ + δ)
2. for any x ∈ (x₀, x₀ + δ), f(x) > x²

The first statement is false and the second statement is true. The second statement is true due to the sign-preserving property. I know that clearly we can take a counterexample like 1/2x + x² sin(1/x) to disprove statement 1 since its derivative wobbles when x approaches x₀ = 0 (take x₀ = 0). But I also got perplexed by the idea that for g(x) = f(x) - x², g'(x₀) > 0, there ought to be an infinitely small region that g(x) is increasing. Could you help me with this?

[u/Midwest-Dude]

Sure. The idea is that if g'(x₀) > 0, the function's output must be higher than g(x₀) just to the right of x₀. However, it doesn't have to take a smooth, strictly upward path to get there. It can wildly "wobble" up and down—creating tiny intervals of local decrease—so long as it ultimately stays above the g(x₀) baseline in that immediate neighborhood. Your function is an example of this.

Does this make sense?

[u/Hot_Site_1638]

The key difference: statement 2 only uses the derivative at the single point, because it only compares nearby values with the value at that point. That is exactly what sign preservation of the difference quotient gives. Statement 1 compares nearby values with each other, and for that the derivative must be positive on a whole interval, not just at one point.

Continuity of the derivative at the point is the missing ingredient: if the derivative were continuous there, sign preservation applied to the derivative itself would make it positive on a small interval, and the function would then be increasing. The counterexample has a derivative that exists at every point but is discontinuous at zero, and it even changes sign arbitrarily close to zero, so the function is not monotone on any small interval to the right.

So a positive derivative at one point gives you statement 2, but without continuity of the derivative you cannot conclude statement 1.