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Determine if (y=tan x + sec x) has any maximums or minimums for pi/2 < x < pi/2 and justify your answer.
 2 years ago
 2 years ago
Determine if (y=tan x + sec x) has any maximums or minimums for pi/2 < x < pi/2 and justify your answer.
 2 years ago
 2 years ago

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badreferencesBest ResponseYou've already chosen the best response.0
y'=sec(x)^2+sin(x), y'=0 when sec(x)^2=sin(x), solve, profit!
 2 years ago

rachnaBest ResponseYou've already chosen the best response.0
wait how dou get sin(x)??
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.0
Oh, sin(x) was my mistake.
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.0
Apparently, I'm still bad at calculus. XP It should be (sin(x))(cos(x)^2).
 2 years ago

rachnaBest ResponseYou've already chosen the best response.0
i thought derivative of sec x is sec x tan x
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.0
sin(x)/cos(x)^2=tan(x)/cos(x)=tan(x)sec(x)
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.0
Same thing. I just didn't remember it, I had to prove it out.
 2 years ago

rachnaBest ResponseYou've already chosen the best response.0
ok so what i do now after i got the derivative ??
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.0
sec(x)^2+tan(x)sec(x)=y'=0, tan(x)+1=0, tan(x)=1, x=arctan(1)
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.0
Woops, my bad. Wait a sec.
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.0
(sec(x)+tan(x))=0, tan(x)=sec(x)
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.0
there we go. when do tan(x) and sec(x) intersect?
 2 years ago

rachnaBest ResponseYou've already chosen the best response.0
wait a minute i need to look at unit circle, i dont have those memorized
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.0
sin(x)/cos(x)=1/cos(x), sin(x)=1 when is this true?
 2 years ago
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