Let [imath]f:(a,b)\to\mathbb{R}[/imath] be a differentiable function such that [imath]\lim_{x \to a^+} f(x)=+\infty[/imath] and [imath]f'(x) \ne 0[/imath] for each [imath]x\in(a,b)[/imath]. Does this imply that [imath]f[/imath] is strictly decreasing in a right neighborhood of [imath]a[/imath]?
I think it's true. Can someone check if my proof is correct, please?
Proof. Since by hypothesis [imath]f'(x) \ne 0[/imath] for each [imath]x \in(a,b)[/imath], then [imath]f'(x)>0[/imath] or [imath]f'(x)<0[/imath] for each [imath]x \in (a,b)[/imath]. Assume, by contradiction, that there exists [imath]0<r<b[/imath] such that [imath]f'(x)>0[/imath] for each [imath]x\in (a,a+r)[/imath], hence [imath]f[/imath] is strictly increasing in [imath](a,a+r)[/imath]. By hypothesis [imath]f(x)\to+\infty[/imath] as [imath]x \to a^+[/imath], hence there exists [imath]R>0[/imath] such that [imath]f(x)>f(a+r)[/imath] for each [imath]x \in (a,a+R)[/imath]. Letting [imath]\rho=\min\{r,R\}[/imath], the limit hypothesis implies that [imath]f(x)>f(a+r)[/imath] and the fact that [imath]f[/imath] is increasing in [imath](a,a+r)[/imath] implies [imath]f(x)<f(a+\rho)<f(a+r)[/imath]; that is, [imath]f(x)>f(a+r)[/imath] and [imath]f(x)<f(a+r)[/imath], a contradiction. Hence, [imath]f'(x)<0[/imath] in [imath](a,a+r)[/imath]; that is, [imath]f[/imath] is strictly decreasing in a right neighborhood of [imath]a[/imath].
I think it's true. Can someone check if my proof is correct, please?
Proof. Since by hypothesis [imath]f'(x) \ne 0[/imath] for each [imath]x \in(a,b)[/imath], then [imath]f'(x)>0[/imath] or [imath]f'(x)<0[/imath] for each [imath]x \in (a,b)[/imath]. Assume, by contradiction, that there exists [imath]0<r<b[/imath] such that [imath]f'(x)>0[/imath] for each [imath]x\in (a,a+r)[/imath], hence [imath]f[/imath] is strictly increasing in [imath](a,a+r)[/imath]. By hypothesis [imath]f(x)\to+\infty[/imath] as [imath]x \to a^+[/imath], hence there exists [imath]R>0[/imath] such that [imath]f(x)>f(a+r)[/imath] for each [imath]x \in (a,a+R)[/imath]. Letting [imath]\rho=\min\{r,R\}[/imath], the limit hypothesis implies that [imath]f(x)>f(a+r)[/imath] and the fact that [imath]f[/imath] is increasing in [imath](a,a+r)[/imath] implies [imath]f(x)<f(a+\rho)<f(a+r)[/imath]; that is, [imath]f(x)>f(a+r)[/imath] and [imath]f(x)<f(a+r)[/imath], a contradiction. Hence, [imath]f'(x)<0[/imath] in [imath](a,a+r)[/imath]; that is, [imath]f[/imath] is strictly decreasing in a right neighborhood of [imath]a[/imath].
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