Question

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

Answer 1

`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`.