Prove that the equation has at least one solution, in interval $x \in [ \frac{1}{ \sqrt[]{3} }, \sqrt[]{3}]$
$$ \frac{1-2x\arctan(x)}{(1+x^2)^2} = – \frac{\pi\sqrt3}{48}$$
I am stumped by this assignment.
I understand Lagrange theorem.
If the function $f(x)$ is defined for every $x \in [a,b]$ and differentiable for every $x \in (a,b)$, then $ \exists c \in (a,b)$, such that $$\frac{f(b) – f(a)}{b-a}=f'(c)$$
What I tried:
$$f(x)=\frac{1-2x\arctan(x)}{(1+x^2)^2}$$ which is obviously defined and differentiable for $x \in [ \frac{1}{ \sqrt[]{3} }, \sqrt[]{3}]$ and $x \in ( \frac{1}{ \sqrt[]{3} }, \sqrt[]{3})$ respectively.
I plugged the function and values in Lagrange theorem formula and got $$\frac{f(\sqrt3) – f(\frac{1}{\sqrt3})}{\sqrt3-\frac{1}{\sqrt3}}=-\frac{24\sqrt3-3\pi}{96}$$ I hope that I am not wrong.
The equation above gives us the value of $f'(x)$ in some point $c\in (\frac{1}{\sqrt3},\sqrt3)$, but how can that bring us closer to the solution?
Please, give me a detailed answer and thank you for your time.
Best Answer
$$f(x)=\frac{\arctan (x)}{1+x^2}\implies f'(x)=\left(\frac{\arctan (x)}{1+x^2}\right)'= \frac{1-2x\arctan(x)}{(1+x^2)^2}$$
Now you need to apply Lagrange theorem to $f(x)$ in the given interval. Since $f(x)$ is continuous and differentiable in the given interval, \begin{align} &\implies \frac{f(\sqrt 3)-f\left(\frac{1}{\sqrt 3}\right)}{\sqrt 3-\frac{1}{\sqrt 3}}=f'(c)\\ &\implies -\frac{\pi\sqrt3}{48}=f'(c)\\ &\implies \exists c \in \left(\frac {1}{\sqrt 3},\sqrt 3 \right ) \, : \frac{1-2c\arctan(c)}{(1+c^2)^2}= -\frac{\pi\sqrt3}{48}\\ \end{align}