Nickolas Taylor
2022-07-05
Answered

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

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behk0

Answered 2022-07-06
Author has **14** answers

The domain should be $x\in [-1,1]$

We have

$f(x)=\frac{1}{\pi}(\frac{\pi}{2}+\mathrm{arctan}x)+\frac{x+1}{{x}^{2}+2x+10}$

so

${f}^{\prime}(x)=\frac{1}{\pi (1+{x}^{2})}+\frac{9-(x+1{)}^{2}}{({x}^{2}+2x+10{)}^{2}}$

This is positive because of $(x+1{)}^{2}\le 4$

Since f(x) is increasing, the answer is $f(1)={47/52}$

We have

$f(x)=\frac{1}{\pi}(\frac{\pi}{2}+\mathrm{arctan}x)+\frac{x+1}{{x}^{2}+2x+10}$

so

${f}^{\prime}(x)=\frac{1}{\pi (1+{x}^{2})}+\frac{9-(x+1{)}^{2}}{({x}^{2}+2x+10{)}^{2}}$

This is positive because of $(x+1{)}^{2}\le 4$

Since f(x) is increasing, the answer is $f(1)={47/52}$

cooloicons62

Answered 2022-07-07
Author has **4** answers

For $x\in [-1,1]$

${f}^{\prime}(x)=\frac{8-2x-{x}^{2}}{(10+2x+{x}^{2}{)}^{2}}+\frac{1}{\pi (1+{x}^{2})}>0$

So, f is increasing in $[-1,1]$. Thus, the maxima is taken at x=1. Plugging that in, the maxima is

$f(1)=\frac{1}{\pi}(\frac{\pi}{2}+\frac{\pi}{4})+\frac{2}{13}=\frac{3}{4}+\frac{2}{13}=\frac{47}{52}$

${f}^{\prime}(x)=\frac{8-2x-{x}^{2}}{(10+2x+{x}^{2}{)}^{2}}+\frac{1}{\pi (1+{x}^{2})}>0$

So, f is increasing in $[-1,1]$. Thus, the maxima is taken at x=1. Plugging that in, the maxima is

$f(1)=\frac{1}{\pi}(\frac{\pi}{2}+\frac{\pi}{4})+\frac{2}{13}=\frac{3}{4}+\frac{2}{13}=\frac{47}{52}$

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