Answer
$$\eqalign{
& \cr
& {\text{Relative minimum at }}\left( { - 0.8814, - 0.5328} \right) \cr
& {\text{Relative maximum at }}\left( {0.8814,0.5328} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2\tanh x - x \cr
& {\text{*Calculate the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {2\tanh x - x} \right] \cr
& f'\left( x \right) = 2{\operatorname{sech} ^2}x - 1 \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& 2{\operatorname{sech} ^2}x - 1 = 0 \cr
& 2{\operatorname{sech} ^2}x = 1 \cr
& {\operatorname{sech} ^2}x = \frac{1}{2} \cr
& {\text{Solving by using a scientific calculator we obtain:}} \cr
& x \approx - 0.8814{\text{ and }}x \approx 0.8814 \cr
& *{\text{Calculate the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {2{{\operatorname{sech} }^2}x - 1} \right] \cr
& f''\left( x \right) = 4\operatorname{sech} x\left( { - \operatorname{sech} x\tanh x} \right) \cr
& f''\left( x \right) = - 4{\operatorname{sech} ^2}x\tanh x \cr
& \cr
& {\text{*Evaluate the second derivative at }}x = - 0.8814 \cr
& f''\left( { - 0.8814} \right) \approx 1.414 > 0 \cr
& {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr
& f\left( x \right){\text{ has a relative minimum at }}\left( { - 0.8814,f\left( { - 0.8814} \right)} \right) \cr
& f\left( { - 0.8814} \right) = - 0.5328 \cr
& {\text{Relative minimum at }}\left( { - 0.8814, - 0.5328} \right) \cr
& \cr
& {\text{*Evaluate the second derivative at }}x = 1.19967 \cr
& f''\left( {0.8814} \right) \approx - 1.414 < 0 \cr
& {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr
& f\left( x \right){\text{ has a relative maximum at }}\left( {0.8814,f\left( {0.8814} \right)} \right) \cr
& f\left( {0.8814} \right) = 0.5328 \cr
& {\text{Relative maximum at }}\left( {0.8814,0.5328} \right) \cr
& \cr
& {\text{Graph}} \cr} $$
