If f (x) = \left(tan\right)^{- 1} \left(\right. \frac{e^{x}}{x + 1} \left.\right) - \left(tan\right)^{- 1} \left(\right. \frac{x + 1}{e^{x}} \left.\right), x >

Engineering

Mathematics

Monotonicity

Question

If f (x) = $\tan^{- 1} (\frac{e^{x}}{x + 1}) - \tan^{- 1} (\frac{x + 1}{e^{x}})$ , x > – 1 then

range of f (x) is $[0, \frac{π}{2})$ .

f (x) is decreasing for all x > – 1.

f (x) is increasing for all x > – 1.

least value of f(f(x)) is 0.

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$f (x) = \tan^{- 1} (\frac{e^{x}}{x + 1}) - \tan^{- 1} (\frac{x + 1}{e^{x}}) = 2 \tan^{- 1} (\frac{e^{x}}{x + 1}) - \frac{π}{2}$

Let $y = \frac{e^{x}}{x + 1} \Rightarrow \frac{dy}{dx} = \frac{(x + 1) e^{x} - e^{x} \cdot 1}{{(x + 1)}^{2}} \Rightarrow \frac{dy}{dx} = \frac{{xe}^{x}}{{(x + 1)}^{2}}$

for x $\in$ (–1, 0) y is decreasing   ⇒ y $\in$ (1, $\infty$ )
for x $\in$ (0, $\infty$ ) y is increasing   ⇒   y $\in$ (1, $\infty$ )
Now   f (x) = $2 \tan^{- 1} y - \frac{π}{2}$

$= [0, \frac{π}{2}), y \in [1, \infty)$