$\text{f}(\text{x})=\tan^{-1}\text{x}$
Clearly, f(x) is continuous on 0, 1 and derivable on 0, 1
Thus, both the conditions of Lagrange's theorem are satisfied.
Concequently, there exist some $\text{c}\in-3,4$
such that$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}=\frac{\text{f}(1)-\text{f}(0)}{1}$
Now,
$\text{f}(\text{x})=\tan^{-1}\text{x}$
$\text{f}'(\text{x})=\frac{1}{1+\text{x}^2},\text{f}(1)=\frac{\pi}{4},\text{f}(0)=0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow\frac{1}{1+\text{x}^2}=\frac{\pi}{4}-0$
$\Rightarrow49\Big(\frac{\pi}{4}-1\Big)=\text{x}^2$
$\Rightarrow\text{x}=\pm\sqrt{\frac{4-\pi}{\pi}}$
Thus, $\text{c}=\sqrt{\frac{4-\pi}{\pi}}\in(0,1)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
Hence, Lagrange's mean value theorem is verified.
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