Consider the equation below. f(x) = 6 cos^2 x - 12 sin x, 0 lessthanorequalto x
ID: 2862346 • Letter: C
Question
Consider the equation below. f(x) = 6 cos^2 x - 12 sin x, 0 lessthanorequalto x lessthanorequalto 2pi 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. Find the inflection points. Find the interval on which f is concave up. (Enter your answer in interval notation.) (x, y) = ((x, y) = (Find the interval on which f is concave down. (Enter your answer in interval notation.)Explanation / Answer
c) The given equation is f(x)=6*cos^2(x)-12sinx
differentiating f(x) with respect to x
f'(x)=-12cosxsinx-12cosx
differentiating f'(x) with respect to x
f''(x)=-12{cosxcosx+sinx(-sinx)-sinx}
f''(x)=-12{cos2(x)-sin2(x)-sinx}
f''(x)=-12{1-sin^2(x)-sin^2(x)-sinx} since sin^2(x)+cos2(x)=1 then cos2(x)=1- sin^2(x)
f''(x)=-12{1-2sin^2(x)-sinx}
f''(x)=-12{1-2sin^2(x)-sinx}=0
by solving above equation we get x=pi/2,(7pi)/6
when x=pi/2, f(x)=6*cos^2(pi/2)-12sin(pi/2)=-12
inflection is (pi/2,-12)
when x=(7pi)/6,6*cos^2((7pi)/6)-12sin((7pi)/6)=21/2
inflection is ((7pi)/6,21/2)
The inflection points are
(x,y)=(pi/2,-12) for smaller x-value
(x,y)=((7pi)/6,21/2) for larger x-value
f''(x)=-12{1-2sin^2(x)-sinx}
f''(0)=-12{1-2sin^2(0)-sin0}=-12 then f''(0)<0
f''((2pi)/3)=-12{1-2sin^2((2pi)/3)-sin(2pi)/3}=6+6*sqrt(3)=16.392>0
The given function f(x) is concave up on interval (pi/2,(7pi)/6) since f''((2pi)/3)>0
The given function f(x) is concave down on interval (0,pi/2) and ((7pi)/6,2pi) since f''(0)<0