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

1 Answer
Nov 26, 2016

Answer:

#x=-(5pi)/6,-pi/2,-pi/6,pi/2#

Explanation:

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#.