f'(Ø) = -12sinØ + 6cos2Ø
= 0 for max/min
I will use x instead of Ø , easier to type
12sinx = 6cos 2x
2sinx = cos 2x
but cos 2x = 1 - 2sin^2 x
2sinx = 1 - 2sin^2 x
2sin^2 x + 2sinx - 1 = 0
sinx = (-2 ± √12)/4
= (-1 ± √3)/2
sinx = (-1+√3)/2 OR sinx = (-1-√3)/2
the last part is not possible since -1<sinx<1
set your calculator to radians to get
x = .3747 or x = π-.3747 = -2.7669
plug those values into the original function to get the max and min values
see:
http://www.wolframalpha.com/input/?i=plot+f%28%CE%B8%29+%3D+12+cos+%CE%B8+%2B+6+sin2%CE%B8+
since were solving a sin(x) function equation and the period is 2π, adding or subtracting multiples of 2π yields more answers.
switching back to Ø
Ø = .3747 + 2nπ OR Ø = -2.7669 + 2nπ , where n is any integer
will yield
Some people also consider the x-intercepts as "critical values"
12 cos θ + 6 sin2θ = 0
2cosØ + sin 2Ø = 0
2cosØ + 2sinØcosØ = 0
cosØ(1 + sinØ) = 0
cosØ = 0 or sinØ = -1
if cosØ =0 , Ø = π/2 or 3π/2
again, the period of cosØ is 2π
so, if cosØ = 0
Ø = π/2 + 2nπ OR Ø = 3π/2 + 2nπ , where n is an integer
if sinØ = -1 , Ø = 3π/2 yielding the same general solution as above
Find the critical numbers of the function. (Use n to denote any arbitrary integer values.)
f(θ) = 12 cos θ + 6 sin2θ
My answer was pi/2 and 3pi/2, but those were wrong. I don't understand the 'n' part, either.
1 answer
1 answer