Find the inflection points?

Question

f(x) = 5cos^2x − 10sinx , 0 ≤ x ≤ 2π 

a) find the inflection points 

Answer 1

The function f(x) = 5cos²x - 10sinx

Differentiate with respect to x

f'(x) = 10cosx(- sinx) - 10cosx

f'(x) = - 10cosx(sinx + 1)

Again differentiate with respect to x.

f''(x) = - 10[cosx(cosx) + (sinx + 1)( - sinx)]

f''(x) = - 10[cos²x - sin²x - sinx]

Equate second derivative = 0

- 10[cos²x - sin²x -  sinx] = 0

cos²x - sin²x - sinx = 0

1 - sin²x - sin²x - sinx = 0

- 2sin²x - sinx + 1 = 0

2sin²x + sinx - 1 = 0

2sin²x + 2sinx - sinx - 1 = 0

2sinx(sinx + 1) - 1(sinx + 1) = 0

(sinx + 1)(2sinx - 1) = 0

sinx + 1 = 0 and 2 sinx = 1

sinx = - 1 and sinx = 1/2

Case 1 : Solve sinx = - 1

x = sin^-1[sin(3π/2)]

x = 3π/2

Case 2 :Solve sinx = 1/2

sinx = sin (π/6)

General solution : If sinθ = sinα, then θ = nπ + (- 1)ⁿα, where n is an integer.

θ = x, α = π/6

For n = 0, x = (0)π + (- 1)⁰(π/6) = π/6

For n = 1, x = (1)π + (- 1)¹(π/6) = 5π/6

For n = 2, x = (2)π + (- 1)²(π/6) = 13π/6

The solutions in the interval 0 ≤ x ≤ 2π are x = 3π/2, π/6, 5π/6.

Comments

Contd...

To find inflection points, using the x - values find the corresponding y - value with the curve.

y = 5cos²x - 10sinx

At x = π/6

y = 5cos²(π/6) - 10sin(π/6) = 5(√3/2)² - 10(1/2) = (15/4) - 5

y = - 5/4

At x = 5π/6

y = 5cos²(5π/6) - 10sin(5π/6) = 5(-√3/2)² - 10(1/2) = (15/4) - 5

y = - 5/4

At x = 3π/2

y = 5cos²(3π/2) - 10sin(3π/2) = 0 - (- 1)

y = 1

Interval    Test Value                      Sign of f''(x)                             Conclusion

(-∞,π/6)     x = 0      f''(0)= -10[cos²(0)-sin²(0)-sin(0)]= -10 < 0     Concave downward.

(π/6, 5π/6)  x = π/2  f''(π/2)=-10[cos²(π/2)-sin²(π/2)-sin(π/2)]=20>0  Concave upward.

(5π/6, 3π/2) x = π      f''(π)=-10[cos²(π)-sin²(π)-sin(π)]=-10 < 0      Concave downward.

(3π/2,∞)     x = 2π  f''(2π)-10[cos²(2π)-sin²(2π)-sin(2π)]=-10 < 0    Concave downward.

There is no change in concavity.Consequently (3π/2, 1) is not inflection point.

Inflection points are (x, y) = (π/6, -5/4) and (5π/6, -5/4).