consider f(x)=sinx+cosx (o,2pi)

Question

find the open intervals on which the function is increasing or decreasing

applye the first derivative test to ideantify relative extrema

Answer 1

The function f(x) = sinx + cosx

Apply derivative on each side with respect of x.

f '(x) = cosx  - sinx

To find relative extrema f '(x) = 0

cosx  - sinx = 0

cosx = sinx

Dvide each side by cosx

(cosx)/(cosx) = (sinx)/(sinx)

1 = tanx

tanx = 1

tan(x) = tan(π/4)

Principal value x = π/4

The genaral solution of tan(x) = tan(α) is x = nπ + α, where n is an integer.

For n = 0 , x = 0π + π/4 = π/4

For n = 1, x = π + π/4 = 5π/4

For n = 2, x = 2π + π/4 = 9π/4

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

y = sinx cosx + 5

For x = π/4

y = sin(π/4) + cos(π/4)

y = (1/√2)+ (1/√2)

y = 2/√2

y = √2

(x,y) = (π/4, √2)

Answer 2

Contd..

(x,y) = (π/4, √2)

For x = 5π/4

y = sin(5π/4) + cos(5π/4)

y = (-√2/2 ) + (-√2/2 )

y = -2√2/2

y = -√2

(x,y) = (5π/4, -√2)

a)The function increasing on the interval (0, π/4) (5π/4, 7π/4)

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

b) Relative maximum point is  (π/4, √2).

Relative minimum point is  (5π/4, -√2).