Answer
$$\eqalign{
& {\text{Concave upward}}:{\text{ }}\left( { - \frac{\pi }{2},0} \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( {0,\frac{\pi }{2}} \right) \cr} $$
Work Step by Step
$$\eqalign{
& y = 2x - \tan x,{\text{ }}\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right) \cr
& {\text{Calculate the first and second derivatives}} \cr
& y' = \frac{d}{{dx}}\left[ {2x - \tan x} \right] \cr
& y' = 2 - {\sec ^2}x \cr
& y'' = \frac{d}{{dx}}\left[ {2 - {{\sec }^2}x} \right] \cr
& y'' = - 2\sec x\frac{d}{{dx}}\left[ {\sec x} \right] \cr
& y'' = - 2\sec x\left( {\sec x\tan x} \right) \cr
& y'' = - 2{\sec ^2}x\tan x \cr
& {\text{Set the second derivative to }}0 \cr
& - 2{\sec ^2}x\tan x = 0 \cr
& - 2\left( {\frac{1}{{{{\cos }^2}x}}} \right)\left( {\frac{{\sin x}}{{\cos x}}} \right) = 0 \cr
& - \frac{{2\sin x}}{{{{\cos }^3}x}} = 0 \cr
& - 2\sin x = 0 \cr
& \sin x = 0 \cr
& {\text{For the interval }}\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right){\text{ }}\sin x = 0{\text{ at }}x = 0 \cr
& {\text{We obtain the critical point }}x = 0 \cr
& {\text{Set the intervals }}\left( { - \frac{\pi }{2},0} \right),\left( {0,\frac{\pi }{2}} \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \frac{\pi }{2},0} \right)}&{\left( {0,\frac{\pi }{2}} \right)} \\
{{\text{Test Value}}}&{ - \frac{\pi }{4}}&{\frac{\pi }{4}} \\
{{\text{Sign of }}f''\left( x \right)}&{f''\left( { - \frac{\pi }{4}} \right) = 4 > 0}&{f''\left( {\frac{\pi }{4}} \right) = - 4 < 0} \\
{{\text{Conclusion}}}&{{\text{Concave upward}}}&{{\text{Concave downward}}}
\end{array}}\]
$$\eqalign{
& {\text{Concave upward}}:{\text{ }}\left( { - \frac{\pi }{2},0} \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( {0,\frac{\pi }{2}} \right) \cr} $$
