we have

y = tanx – 2x

y’ = sec²x – 2

y’’ = 2sec²xtanx

For maxima and minima,

y’ = 0

sec2x = 2

secx = ±√2

x =

y’’() = 4 > 0

x = is the point of minima

y’’() = – 4 < 0

x = is the point of maxima

hence

max value = f() = – 1 –

min value = f() = 1 –