Question

By using $LMVT$, prove that
$\frac{\beta -\alpha }{1+{\beta }^2}<{\tan }^{-1}\beta -{\tan }^{-1}\alpha <\frac{\beta -\alpha }{1+{\alpha }^2},\beta -\alpha <0$.

Answer 1

$\begin{array}{c}letf\left(x\right)={\tan }^{-1}x\\ f^{\prime }\left(x\right)=\frac{1}{1+x^2}\\ ApplyingMeanValuetheoremonf\left(x\right)is\\ \alpha <x<\beta \frac{f\left(\beta \right)-f\left(\alpha \right)}{\beta -\alpha }=f^{\prime }\left(C\right)=\frac{1}{1+x^2}where\alpha <C<\beta \\ Now\\ f^{\prime \prime }\left(x\right)=-\frac{1}{\left(1+x^2\right)}2x<0whenx>0\\ \Rightarrow f^{\prime }\left(x\right)isdecrea\sin gforx>0\\ Hence\\ \alpha <C<\beta \\ \Rightarrow f^{\prime }\left(\alpha \right)>f^{\prime }\left(C\right)>f^{\prime }\left(\beta \right)\\ \Rightarrow \frac{1}{1+{\alpha }^2}>\frac{f\left(\beta \right)-f\left(\alpha \right)}{\beta -\alpha }>\frac{1}{1+{\beta }^2}\\ \Rightarrow \frac{\beta -\alpha }{1+{\alpha }^2}>{\tan }^{-1}\beta -{\tan }^{-1}\alpha >\frac{\beta -\alpha }{1+{\beta }^2}\end{array}$