Critical numbers of trig functions

jeonw

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Oct 10, 2010
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2
1. The problem statement, all variables and given/known data
Find the critical Numbers

f(theta)=2sec(theta)+tan(theta)

2. The attempt at a solution
I found the derivative and set it equal to zero and to reduce writing I substituted x for theta
f'(x)=2sec(x)tan(x)+sec^2(x)
factored out the sec(x)
sec(x)[2tan(x)+sec(x)]=0

My question is where are the critical numbers? do the critical numbers exist where sec(x) is underfined because sec(x) will never equal 0.
 
Hello, jeonw!

You're doing fine . . .


1. Find the critical numbers of: .f(x)=2secx+tanx\displaystyle f(x) \:=\:2\sec x + \tan x


The attempt at a solution:

I found the derivative and set it equal to zero: .f(x)=2secxtanx+sec2x=0\displaystyle f'(x)\:=\:2\sec x\tan x+\sec^2x \:=\:0

Factored: .secx(2tanx+secx)=0\displaystyle \sec x (2\tan x +\sec x ) \:=\:0

Set each factor equal to zero and solve.


\(\displaystyle \sec x \:=\:0\;\hdots \text{ impossible.}\)


2tanx+secx=02sinxcosx+1cosx=02sinx+1cosx=0\displaystyle 2\tan x + \sec x \:=\:0 \quad\Rightarrow\quad \frac{2\sin x}{\cos x} + \frac{1}{\cos x} \:=\:0 \quad\Rightarrow\quad \frac{2\sin x +1}{\cos x}\:=\:0

. . 2sinx+1=0sinx=12\displaystyle 2\sin x + 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:-\tfrac{1}{2}


\(\displaystyle \text{Therefore: }\;x \;=\;\begin{Bmatrix}\text{-}\dfrac{\pi}{6} + 2\pi n \\ \\[-3mm] \dfrac{7\pi}{6} + 2\pi n \end{Bmatrix}\;\text{ for any integer }n.\)

 
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