How to find the maximum value of $x\cos^{-1}\left(x\right)$? I wanted to find the maximum value of $x\cos^{-1}\left(x\right)$ through differentiation, but upon on differentiation I get $$f'(x) = \arccos\left(x\right) - \frac{x}{\sqrt{1-x^2}}$$  to find the maximum I equated the function to zero:    $$\arccos\left(x\right) - \frac{x}{\sqrt{1-x^2}} = 0$$
But I am unable to find the roots of this equation. Could someone show how to find the zeros for this equation?
 A: As you wrote
$$f(x)=x \cos ^{-1}(x)\qquad \text{and} \qquad f'(x)=\cos ^{-1}(x)-\frac{x}{\sqrt{1-x^2}}$$
The problem would be simple using Newton method since, from the graph of $f'(x)$, you probably notice that the solution is close to $x=0.6$.
Without numerical method, since $x=\frac 12$ is a nice number for the arcosine, perform a series expansion. Thsi would give
$$f'(x)=\left(\frac{\pi }{3}-\frac{1}{\sqrt{3}}\right)-\frac{14
   \left(x-\frac{1}{2}\right)}{3 \sqrt{3}}-\frac{10 \left(x-\frac{1}{2}\right)^2}{3
   \sqrt{3}}-\frac{152 \left(x-\frac{1}{2}\right)^3}{27 \sqrt{3}}-\frac{724
   \left(x-\frac{1}{2}\right)^4}{81
   \sqrt{3}}+O\left(\left(x-\frac{1}{2}\right)^5\right)$$ Now, perform a series reversion and obtain
$$x=\frac{1}{2}-t-\frac{5 t^2}{7}+\frac{82 t^3}{441}+\frac{5287
   t^4}{9261}+O\left(t^5\right)\quad \text{with}\quad t=\frac{3\sqrt{3}}{14}  \left(f'(x)-\frac{\pi }{3}+\frac{1}{\sqrt{3}}\right)$$ and we want $f'x)=0$.
So, this truncated expansion gives as an estimate
$$x=\frac{10117671+3552168 \sqrt{3} \pi -264690 \pi ^2-29184 \sqrt{3} \pi ^3+5287 \pi^4}{39530064}$$ which, numerically, is $0.652206$ quite close to the value @Mariusz Iwaniuk gave in the first comment . For sure, adding a few more terms, the estimate will be closer and closer to the solution.
For example, adding the next term in the initial expansion, the same procedure would give
$$x=\frac{354165816+124246995 \sqrt{3} \pi -9106380 \pi ^2-1074030 \sqrt{3} \pi
   ^3+211340 \pi ^4-1753 \sqrt{3} \pi ^5}{1383552240}$$ which is $0.652194$.
A: \begin{align*}
f(x) &=xcos^{-1}(x) \\
f'(x) &=cos^{-1}(x) - \frac{x}{\sqrt{1-x^{2}}} \\
0 &=cos^{-1}(x) - \frac{x}{\sqrt{1-x^{2}}} \\
\frac{x}{\sqrt{1-x^{2}}} &=cos^{-1}(x) \\
cos(\frac{x}{\sqrt{1-x^{2}}}) &= x \\
\end{align*}
Now substitute $x = cos\alpha$:
\begin{align*}
cos(\frac{cos\alpha}{\sqrt{1-(cos\alpha)^{2}}}) &= cos\alpha \\
cos(\frac{cos\alpha}{sin\alpha}) &= cos\alpha \\
\frac{cos\alpha}{sin\alpha} &= \alpha \\
cot\alpha &= \alpha \\
\end{align*}
I will admit that I do not know how to analytically find the $\alpha$ that makes this true (I looked online and the only answers I found were using Newtons method or other ways of approximating the answer). Either way, the cosine of this $\alpha$ is the maximum point of $f(x)$.
