# Critical points of Sec x on an interval?

g(x)= sec(x) on the closed interval [ -pi/6, pi/3]..... I got the derivative of Sec x Tan x but I don't know what to do from there

Relevance
• 8 years ago

g(x) =  y = sec(x)

dy ⁄ dx = sec(x) • tan(x) ... set to zero for min/max

0 = sec(x) • tan(x)

... either or both of the terms: sec(x) and tan(x) must be zero.

First Term Solution:

sec(x) = 0

1 ⁄ cos(x) = 0

... no solution since cos(x) never equals infinity

Second Term Solution:

tan(x) = 0

sin(x) ⁄ cos(x) = 0 ... solution is when numerator is zero

sin(x) = 0

x = 0 ... on the interval: [- π ⁄ 6, π ⁄ 3 ]

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2nd derivative:

d²y ⁄ dx²  =  sec(x) • sec²(x)  +  tan(x) • sec(x) • tan(x)

d²y ⁄ dx²  =  sec³(x)  +  tan²(x) • sec(x)

d²y ⁄ dx²  =  sec(x) • [ sec²(x)  +  tan²(x) ]

at x = 0 :

d²y ⁄ dx²  =  sec(0) • [ sec²(0)  +  tan²(0) ]

d²y ⁄ dx²  =  1 • [ 1  +  0 ]

d²y ⁄ dx²  =  1 ... which is > 0 , so at (x=1) there is a minimum

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... at the boundaries of the interval: [- π ⁄ 6, π ⁄ 3 ]

... when checking the polarity of the 2nd derivative ONLY,

... the bracketed multiplier: [ sec²(- π ⁄ 6)  +  tan²(- π ⁄ 6) ]

... can be ignored since it is always > 0

at x = - π ⁄ 6 : Polarity Check

d²y ⁄ dx²  =  sec(- π ⁄ 6) • [ sec²(- π ⁄ 6)  +  tan²(- π ⁄ 6) ]

d²y ⁄ dx²  =  sec(- π ⁄ 6) ... ignoring the bracketed multiplier

= which is: > 0

at x = π ⁄ 3 : Polarity Check

d²y ⁄ dx²  =  sec(π ⁄ 3) • [ sec²(π ⁄ 3)  +  tan²(π ⁄ 3) ]

d²y ⁄ dx²  =  sec(π ⁄ 3) ... ignoring the bracketed multiplier

= which is: > 0

In fact: sec(x) is ALWAYS > 0 , on the interval : [- π ⁄ 6, π ⁄ 3 ]

... so the 2nd derivative doesn't change sign ...

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The only critical point on the given interval is a minimum at x = 0, g(x) = 1