An analyst claims that the function has a local maximum at some point . Evaluate this claim using…

The analyst's claim is incorrect. For all real $x$, $x^2 \ge 0$, so $\sqrt{1+x^2} \ge 1$ and $1+x^2 \ge 1$. Thus, $\frac{2}{\sqrt{1+x^2}} > 0$ and $\frac{2}{1+x^2} > 0$, which means $f'(x) > 0$ for all $x \in \mathbb{R}$. Since $f'(x)$ is strictly positive everywhere, $f(x)$ is strictly increasing on $\mathbb{R}$ and $f'(x)$ never equals zero. Therefore, $f(x)$ has no local extrema (neither maximum nor minimum).