Homework Help

f(x)=sin(2x)-(3/4)tan(x) how can I educate the sign of f(x) using derivation ?

user profile pic

pourjour | Student, Undergraduate | eNoter

Posted May 21, 2012 at 5:59 PM via web

dislike 1 like

f(x)=sin(2x)-(3/4)tan(x) how can I educate the sign of f(x) using derivation ?

2 Answers | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 22, 2012 at 5:05 PM (Answer #1)

dislike 2 like

You need to use derivative of the function to find the intervals of monotony of function such that:

`f'(x) = 2cos(2x) - 3/(4cos^2 x)`

Solving the equation f'(x) = 0 yields:

`2cos(2x) - 3/(4cos^2 x) = 0`

You need to substitute `2cos^2 x - 1`  for `cos 2x`  such that:

`2(2cos^2 x - 1) - 3/(4cos^2 x) = 0`

`4cos^2 x - 2 - 3/(4cos^2 x) = 0`

`16cos^4 x - 8cos^2 x - 3 = 0`

You should come up with the substitution `cos^2 x =y`  such that:

`16y^2 - 8y - 3 = 0`

`y_(1,2) = (8+-sqrt(64 + 192))/32`

`y_(1,2) = (8+-sqrt256)/32`

`y_(1,2) = (8+-16)/32`

`y_1 = 24/32 =gt y_1 = 3/4`

`y_2 = -8/32 =gt y_2 = -1/4`

You need to solve for x the equations `cos^2 x = y_(1,2)`

`cos^2 x = 3/4 =gt cos x = +-sqrt3/2`

`x = pi/6`  or x = `11pi/6` ``

`x = 5pi/6 or x = 7pi/6`

`cos^2 x = -1/4`  impossible

Hence, the derivative is positive if x

`16cos^4 (2pi/3) - 8cos^2(2pi/3) - 3 = 0`

`16/16 - 8/4 - 3 = 0`

`-2 - 2 = -4 lt 0`

Notice that if `x in [0,pi/6), x in (5pi/6,pi], x in (11pi/6,2pi], ` the derivative is positive, hence the function increases and is positive.

user profile pic

pourjour | Student, Undergraduate | eNoter

Posted May 21, 2012 at 8:08 PM (Answer #2)

dislike 0 like
Or whatever technique

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes