Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = 9 cos2(x) − 18 sin(x), 0 ≤ x ≤ 2𝜋

Question

Consider the equation below. (If an answer does not exist, enter DNE.)

f(x) = 9 cos²(x) − 18 sin(x),      0 ≤ x ≤ 2𝜋

(a) Find the inflection points.

(x, y) = (smaller x-value)

(x, y) = (larger x-value)

(b)

Find the interval on which f is concave up. (Enter your answer using interval notation.)

(c)

Find the interval on which f is concave down. (Enter your answer using interval notation.)

Answer 1

f(x)=9cos²(x) - 18sin(x)

f'(x) = 18cos(x)(-sin(x)) - 18cos(x)= -18cos(x)[sin(x) + 1] ; (critical numbers π/2 and 3π/2)

f''(x) = -18cos(x)[cos(x)] + [sin(x) + 1][-18(-sin(x))]

= -18cos²(x) +18sin²(x) + 18sin(x)

= 18sin²(x) - 18cos²(x) + 18sin(x)

When does f'' = 0?

sin²(x) - cos²(x) + sin(x) = 0 (divide all terms by 18)

sin²(x) - (1 - sin²(x)) + sin(x) = 0

2sin²(x) + sin(x) - 1 = 0

[2sin(x) - 1] [sin(x) + 1] = 0

sin(x) = 1/2 or sin(x) = -1

x = sin^-1(1/2) or x = sin^-1(-1)

x = π/6, 5π/6, 3π/2

Determine sign of 2nd derivative for each interval created by those hypercritical numbers:

f''(0) < 0, so f is concave down between 0 and π/6.

f''(π/2) > 0, so f is concave up between π/6 and 5π/6.

Inflection point when x = π/6.

f''(π) < 0, so concave down between 5π/6 and 3π/2.

Concavity changes at 5π/6 so inflection point there.

f''(2π) < 0, so concave down from 3π/2 to 2π. Concavity did not change at 3π/2.

Points of inflection at:

(π/6, f(π/6)) and (5π/6. f(5π/6))

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