What are the critical values, if any, of #f(x)=2secx + tan x in[-pi,pi]#?

Question

What are the critical values, if any, of #f(x)=2secx + tan x  in[-pi,pi]#?

Emilia Aguilar · Accepted Answer

#x=-(5pi)/6,-pi/2,-pi/6,pi/2# Quickly showing the derivations for the derivatives of #secx=1/cosx# and #tanx=sinx/cosx#: #d/dxsecx=d/dx(1/cosx)# #color(white)(d/dxsecx)=d/dx(cosx)^-1# #color(white)(d/dxsecx)=-(cosx)^-2d/dx(cosx)# #color(white)(d/dxsecx)=-1/cos^2x(-sinx)# #color(white)(d/dxsecx)=1/cosx(sinx/cosx)# #color(white)(d/dxsecx)=secxtanx# #d/dxtanx=d/dx(sinx/cosx)# #color(white)(d/dxtanx)=((d/dxsinx)cosx-sinx(d/dxcosx))/cos^2x# #color(white)(d/dxtanx)=(cosx(cosx)-sinx(-sinx))/cos^2x# #color(white)(d/dxtanx)=(cos^2x+sin^2x)/cos^2x# #color(white)(d/dxtanx)=1/cos^2x# #color(white)(d/dxtanx)=sec^2x# So, if #f(x)=2secx+tanx#, then: #f'(x)=2secxtanx+sec^2x# #color(white)(f'(x))=(2sinx)/cos^2x+1/cos^2x# #color(white)(f'(x))=(2sinx+1)/cos^2x# A critical value will occur when #f'(x)=0# or when #f'# is undefined. We note that #f'(x)=0# when #2sinx+1=0#. This means that #sinx=-1/2#. On the interval #x in[-pi,pi]# this means there are critical values at #color(blue)(x=-pi/6,-(5pi)/6#. #f'# is undefined when #cos^2x=0#, so #cosx=0#. In the interval this is at #color(blue)(x=-pi/2,pi/2#.

Emma Aldridge · Answer

undefined The critical values of $ f(x) = 2\sec x + \tan x $ in the interval $[-π, π]$ are $ x = -π/3 $ and $ x = π/3 $.