# Prove there exists some $x_0$ for a differentiable function

A real-valued function $$f$$ is defined and differentiable on $$[a,b]$$ ($$b-a\geq{4}$$). Prove that there exists $$x_0 \in (a,b)$$ for which $$f'(x_0)<1+f^2(x_0)$$

On the one hand, the statement resembles very much the classical theorem by Lagrange according to which there exists some $$\varepsilon$$ for which $$f'(\varepsilon)(b-a)=f(b)-f(a)$$

Nevertheless, what we have here is an inequality, which is more tricky. The limitation ($$b-a\geq{4}$$) makes me think that using trigonometry, in this case, might be a good approach, but I have no idea how exactly that could be used.

Thanks in advance for any hints.

## 3 Answers

If this is not true we would get $$\frac{f’(x)}{1+f^2(x)}\ge 1$$ for all $$x$$ in $$[a,b]$$. Integrating we get $$\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)>\arctan(f(b))-\arctan(f(a))\ge b-a$$ That is $$b-a<\pi<4$$, so, if $$b-a>4$$ (or even $$b-a\ge \pi$$) then there must be some $$x_0\in [a,b]$$ such that $$f’(x_0)<1+f^2(x_0)$$.

You can also go a more direct way after recognizing the connection to the tangent function: Consider the function $$g(x)=\arctan(f(x)),$$ then by the properties of the inverse tangent function, the interval length bound and the mean value theorem there exists some $$x_0\in(a,b)$$ so that $$\frac{\pi}4\ge\frac{|g(b)-g(a)|}{b-a}=|g'(x_0)|=\frac{|f'(x_0)|}{1+f(x_0)^2}$$

• Simple and elegant, thank you. Glad to see the mean value theorem being used – Don Draper Jun 11 at 19:34

Clearly scaling the function vertically or horizontally does not affect the question, so assume $$a=0, b>4$$, and WLOG $$f(a)=0$$. Let $$y=f(x)$$ and suppose for contradiction that $$\frac{dy}{dx} \ge 1+y^2$$ for all $$x$$ in the interval. Then we have $$\frac{dx}{dy} \le \frac{1}{1+y^2}$$, and integrating from gives $$x \le \tan^{-1}{y}+c$$. So $$y>\tan{x-c}$$. But $$x$$ varies over an interval of width at least $$4\ge \pi$$, so there exists some point where $$\tan{x-c}$$ approaches infinity, and then $$y$$, being continuous, must also approach infinity and become undefined at the asymptote. This is a contradiction, as desired.

• Actually, the main idea is to show that $y=f(x)$ can get very large and can't be less or equal to $y'$ for all $x$? I don't quite understand the "asymptote" part. Why do we need to mention it at all? – Don Draper Jun 11 at 17:57
• Why is setting $f(a)=0$ "WLOG"? – LutzL Jun 11 at 18:25