Consider the equation below. f(x) = 9 cos^2x - 18 sin x, 0 lessthanorequalto x l
ID: 2875298 • Letter: C
Question
Consider the equation below. f(x) = 9 cos^2x - 18 sin x, 0 lessthanorequalto x lessthanorequalto 2 pi Find the interval on which f is increasing. (Enter your answer in interval notation.) Find the interval on which f is decreasing. (Enter your answer in interval notation.) Find the local minimum and maximum values of f. local minimum local maximum Find the inflection points. (x, y) = (smaller x-value) ) (x, y) = (larger x-value) Find the interval on which f is concave up. (Enter your answer in interval notation.) Find the interval on which f is concave down. (Enter your answer in interval notation.)Explanation / Answer
Given f(x)=9cos^2(x)-18sinx
=> f ' (x) = 18cos(x)(-sin(x)) -18cos(x) = - 18cos(x)[sin(x) +1]
f '' (x) = 18sin(x)[sin(x)+1] - 18cos(x)cos(x) = 18[ sin^2(x) - cos^2(x) +sin(x)) =18(2sin^2(x)+sin(x)-1)
Where I used cos^2(x)=1- sin^2(x) to simplify the second derivative.
f is increasing if f ' >0 and decreasing if f ' < 0. To find where it changes signs, we can see where it is equal to zero.
Setting f ' = - 18cos(x)[sin(x) +1] =0 if cos(x) = 0 => x=npi/2 => x=pi/2, 3pi/2
or sin(x) = -1 => x=3pi/2.
On [0,2pi], the derivative is only zero once at pi/2 which is approximately 1.57. Checking the sign to the left and right, it is negative to the left and positive to the right. So it is decreasing on (0, pi/2) U(3pi/2,2pi) and increasing on (pi/2,3pi/2) (3pi/2,2pi).)
(b) Note that its decreasing as it approaches pi/2, stops decreasing and starts increasing. This tells you that it must have reached a local minimum at pi/2. A similar analysis tells you that is must reach a local maximum at 3pi/2.
f(pi/2)=-18 minimum
f(3pi/2)=18 maximum
(c) Inflection points occur where the function is differentiable, the second derivative is zero, and the second derivative actually changes signs about the point. The second derivative is zero if 2sin^2(x)+sin(x)-1=0
(2sin(x) -1)(sin(x)+1)=0 ==> sin(x) =1/2 or sin(x) = -1
sin(x) = 1/2 if x = pi/6 or 5pi/6 and sin(x) = -1 if x=3pi/2
therfore (x,y)=(pi/6,-9/4)
f(x,y)=(5pi/6,-9/4)
f(x,y)=(3pi/2,18)
(d) f '' is negative on [0, pi/6), positive on (pi/6,5pi/6), negative on (5pi/6,2pi). So f is concave up on (pi/6,5pi/6) and down on (0,pi/6) and (5pi/6,2pi). It has inflection points when x=pi/6 and x=5pi/6.