$$f'\left( x \right) = \sqrt x \sin x$$

At local maxima or minima, $$f'\left( x \right) = 0$$

$$ \Rightarrow x = 0$$ or $$sin$$ $$x=0$$

$$ \Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)$$

$$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$$

$$ = {1 \over {2\sqrt x }}\left( {2x\,\cos \,x + \sin \,x} \right)$$

At $$x = \pi ,$$ $$f''\left( x \right) < 0$$

Hence, local maxima at $$x = \pi $$

At $$x = 2\pi ,\,\,\,f''\left( x \right) > 0$$

Hence local minima at $$x = 2\pi $$

$$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \over {{e^{2x}} + 2}}$$

$$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$$

$$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$$

$${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2 $$

maximum $$f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$$

$$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$$

Since $$0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow $$ for some $$c \in R$$

$$f\left( c \right) = {1 \over 3}$$

Since tangent is parallel to $$x$$-axis,

$$\therefore$$ $${{dy} \over {dx}} = 0 \Rightarrow 1 - {8 \over {{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3$$

Equation of tangent is $$y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3$$

$$f\left( x \right) = \left\{ {\matrix{ {k - 2x,\,\,\,\,if\,\,\,\,x \le - 1} \cr {2x + 3,\,\,\,\,if\,\,\,\,x > - 1} \cr } } \right.$$



This is true where $$k=-1$$