Question

-2(secx)^2Tanx=0 Solve for x in the interval ((-pi/2),(pi/2))

Answer 1

The original q was "Determine the open intervals on which the function f(x)=2x-tanx,(-pi/2,pi/2) is concave upward or downwards"

Answer 2

I foudn the 2nd derivative but Idk what x's make this 0

Answer 3

-2(secx)^2Tanx=0
so,
sec x =0 or tan x =0

Answer 4

Yeah, I don't think i paid attention well enough when the teacher did all the trig stuff. idk how to find those without looking at a chart. I have one but can u teach me how it actually works?

Answer 5

tan x = 0 for what all values of x
so sin x/ cos x = 0
so sin x = 0
for which values ?

Answer 6

x=0
??

Answer 7

yes, in interval -pi/2 to pi/2, it will be x=0 only
and for what vaules will be sec x = 0 ?

Answer 8

secx=1/cosx... how do I get 0 with that? 0_0

Answer 9

thats because, sec x can never be 0, so u have only one solution, x=0

Answer 10

so for -2(secx)^2Tanx=0
x=0 only , in -pi/2, pi/2

Answer 11

ohhh! Ok! Thank you!

Answer 12

welcome ^_^

Answer 13

:D