Question

If n < 0 and f(x) = e^(nx) + e^(- nx) is monotonically decreasing. Find the interval to which x belongs.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = {e}^{nx} + {e}^{ - nx} \: where \: n < 0$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx} \bigg[{e}^{nx} + {e}^{ - nx}\bigg]$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {e}^{x} = {e}^{x} \: }}$

So, using this, we get

$\rm :\longmapsto\:f'(x) = {e}^{nx}(n) + {e}^{ - nx}( - n)$

$\rm :\longmapsto\:f'(x) = n{e}^{nx} - n {e}^{ - nx}$

$\rm :\longmapsto\:f'(x) = n\bigg[{e}^{nx} - {e}^{ - nx}\bigg]$

$\rm :\longmapsto\:f'(x) = n\bigg[{e}^{nx} - \dfrac{1}{{e}^{nx}} \bigg]$

$\rm :\longmapsto\:f'(x) = n\bigg[ \dfrac{{e}^{2nx} - 1}{{e}^{nx}} \bigg]$

Now, It is given that f(x) is monotonically decreasing.

$\rm \implies\:f'(x) < 0$

$\rm :\longmapsto\: n\bigg[ \dfrac{{e}^{2nx} - 1}{{e}^{nx}} \bigg] < 0$

As n < 0 and exponential functions are always positive.

$\rm \implies\:{e}^{2nx} - 1 > 0$

$\rm \implies\:{e}^{2nx} > 1$

$\rm \implies\:{e}^{2nx} > {e}^{0}$

$\rm \implies\:2nx > 0$

As n < 0

$\rm \implies\:x < 0$

$\bf \implies\:x \: \in \: ( - \infty , \: 0)$

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More to know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$