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

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find the open intervals where function is increasing or decreasing

apply 1st derivative test to find all relative extrema
asked Oct 2, 2014 in CALCULUS by anonymous

2 Answers

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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) = (/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) = (/4, 5.5)

 

answered Oct 2, 2014 by david Expert
0 votes

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) = (/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)

 

answered Oct 2, 2014 by david Expert

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