Nickolas Taylor

2022-07-05

If $f\left(x\right)=\frac{1}{\pi }\left(\mathrm{arcsin}x+\mathrm{arccos}x+\mathrm{arctan}x\right)+\frac{x+1}{{x}^{2}+2x+10}\phantom{\rule{thickmathspace}{0ex}},$, Then max value of f(x)

behk0

The domain should be $x\in \left[-1,1\right]$
We have
$f\left(x\right)=\frac{1}{\pi }\left(\frac{\pi }{2}+\mathrm{arctan}x\right)+\frac{x+1}{{x}^{2}+2x+10}$
so
${f}^{\prime }\left(x\right)=\frac{1}{\pi \left(1+{x}^{2}\right)}+\frac{9-\left(x+1{\right)}^{2}}{\left({x}^{2}+2x+10{\right)}^{2}}$
This is positive because of $\left(x+1{\right)}^{2}\le 4$
Since f(x) is increasing, the answer is $f\left(1\right)=47/52$

cooloicons62

For $x\in \left[-1,1\right]$
${f}^{\prime }\left(x\right)=\frac{8-2x-{x}^{2}}{\left(10+2x+{x}^{2}{\right)}^{2}}+\frac{1}{\pi \left(1+{x}^{2}\right)}>0$
So, f is increasing in $\left[-1,1\right]$. Thus, the maxima is taken at x=1. Plugging that in, the maxima is
$f\left(1\right)=\frac{1}{\pi }\left(\frac{\pi }{2}+\frac{\pi }{4}\right)+\frac{2}{13}=\frac{3}{4}+\frac{2}{13}=\frac{47}{52}$

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