Consider the equation below. f(x)= 7cos^ 2x-14 sin x, 0<_x <_ 2pi (a) Find the i
ID: 2838528 • Letter: C
Question
Consider the equation below. f(x)= 7cos^2x-14 sin x, 0<_x <_ 2pi (a) 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.) (b) Find the local minimum and maximum values of f. local minimum local maximum (c) 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
f(x) = 7cos2x - 14 sinx
f'(x) = -14sinx*cosx - 14cosx
f'(x) = -14cosx(sinx + 1)..........................(1)
f''(x) = 14sinx(sinx + 1) - 14cosx * cosx
f''(x) = 14sin2x - 14cos2x + 14sinx
f''(x) = 14sin2x - 14(1 - sin2x) + 14sinx
f''(x) = 14sin2x - 14 + 14sin2x + 14sinx
f''(x) = 14(2sin2x + sinx - 1)
f''(x) = 14(2sinx - 1)(sinx + 1)...........(2)
(a)
For increasing, f'(x) > 0
14cosx(sinx + 1) > 0
14cosx(sinx + 1) < 0
sinx + 1 can never be negative, so cosx < 0
x = (pi/2, 3pi/2).......increasing
(b)
for decreasing. f'(x) < 0
x = (0, pi/2) U (3pi/2, 2pi).........decreasing
(c)
for inflection points, f''(x) = 0
14(2sinx - 1)(sinx + 1) = 0
2sinx - 1 = 0, sinx + 1 = 0
sinx = 1/2, -1
x = pi/6, 5pi/6, 3pi/2...........inflection points
f(pi/6) = 7cos2x - 14 sinx = 7*3/4 - 14*1/2 = 21/4 - 7 = -7/4
f(5pi/6) = 7cos2x - 14 sinx = 7*3/4 - 14*1/2 = 21/4 - 7 = -7/4
f(3pi/2) = 7cos2x - 14 sinx = 0 - 14*-1 = 14
For concave up, f''(x) > 0
14(2sinx - 1)(sinx + 1) > 0
since sinx + 1 is always positive
2sinx - 1 > 0
x = (pi/6, 5pi/6).......................concave up
x = (0, pi/6) U (5pi/6, 2pi).........concave down