Prove that $\arctan{x}=\frac{1}{x^2}$ has only one solution on the set of real numbers.

Question

Prove that $\arctan{x}=\frac{1}{x^2}$ has only one solution on the set of real numbers.

I need some help with it, would greatly appreciate it.

Answer 1

For $x > 0$, $\arctan x$ is strictly increasing while $\dfrac{1}{x^2}$ is strictly decreasing.

For $x < 0$, $\arctan x < 0 < \dfrac{1}{x^2}$.

What does this tell you about the number of positive and negative solutions?

Answer 2

Infer from the graph plotting. Draw the graph of $y=1/x^2$ on paper and then make the graph of $\arctan(x)$ wherever they intersect is your solution and number of points of intersection are your number of solution. This is answer is given on the assumption that you know the basic graphs of $1/x^2$ and $\arctan(x)$.