f(x)=sinxcosx+5 on interval (0,2pi)

Question

find the open intervals where function is increasing or decreasing

apply 1st derivative test to find all relative extrema

Answer 1

The function f(x) = sinx cosx + 5

Apply product rule in derivatives d/dx (uv) = uv' + vu'

u = sinx, v = cosx

u' = cosx, v' = - sinx

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

f ' (x) = - sin² x + cos² x

To find relative extrema f ' (x) = 0

- sin² x + cos² x = 0

- sin² x + (1 - sin² x) = 0

- sin² x +1 - sin² x = 0

1 - 2 sin² x  = 0

2 sin² x = 1

sin² x = 1/2

sin² x = [1/√2]²

sin² x = sin²[π/4]

x = nπ + π/4

The general solution of sin²(θ) = sin²(α) is θ = nπ ± α, where n is an integer.

For n = 0 , x = 0π ± π/4 = π/4, -π/4,

For n = 1, x = π ± π/4 = 5π/4, 3π/4

For n = 2, x = 2π ± π/4 = 9π/4, 7π/4

The solutions in the interval (0, 2π) π/4, 3π/4,5π/4,7π/4.

y = sinx cosx + 5

For x = π/4

y = sin(π/4) cos(π/4) + 5

y = (1/√2)(1/√2) + 5

y = 1/2 + 5

y = 11/2 = 5.5

(x, y) = (π/4, 5.5)

For x = 3π/4

y = sin(3π/4) cos(3π/4) + 5

y = (√2/2 )(-√2/2 ) + 5

y = -1/2 + 5

y = 4.5

(x, y) = (3π/4, 4.5)

For x = 5π/4

y = sin(5π/4) cos(5π/4) + 5

y = (-√2/2 )(-√2/2 ) + 5

y = 1/2 + 5

y = 5.5

(x, y) = (5π/4, 5.5)

 

Answer 2

Contd..

For x = 7π/4

y = sin(7π/4) cos(7π/4) + 5

y = (-√2/2 )(√2/2 ) + 5

y = - 1/2 + 5

y = 4.5

(x, y) = (7π/4, 4.5)

a)The function increasing on the interval (0, π/4) (3π/4, 5π/4)

The function is decreasing on the interval (π/4,3π/4) (5π/4,7π/4)

b) Relative maximum points are (π/4, 5.5) and (5π/4, 5.5)

Relative minimum points are (3π/4, 4.5) and (7π/4, 4.5)