global max and min

Question

Find the global maximum and the global minimum of f(x) = 2 sin(x) + cos(2x) on the interval [0, 2].

Show all your calculations.

Answer 1

The function f(x) = 2 sin(x) + cos(2x)

Differentiate with respect to x.

f'(x) = 2 cos(x) - 2sin(2x)

Equate the first derivative = 0

2 cos(x) - 2sin(2x) = 0

2 cos(x) - 4 sin(x) cos(x) = 0

2 cos(x)[ 1 - 2 sinx] = 0

2 cos(x) = 0 and 1 - 2 sin(x) = 0

Solve 2cos(x) = 0

cos(x) = 0

cos(x) = cos(π/2)

Principal value x = π/2

Solve 1 - 2 sin(x) = 0

1 = 2sin(x)

sin(x) = 1/2

sin(x) = sin(π/6)

Principal value π/6.

Solutions are π/2 and π/6.

The critical numbers are x = π/2 and x = π/6. So, the number of critical points is 2.

	x - value	f(x) = 2 sin(x)+cos(2x)	Conclusion
Left Endpoint	0	f(0) = 2sin(0)+cos(2(0))= 1	Minimum
Critical number	π/2	f(π/2) = 2sin(π/2)+cos(2(π/2))= 2 - 1= 1	Minimum
Critical number	π/6	f(π/6) = 2sin(π/6)+cos(2(π/6))=1+1/2=3/2	Maximum
Right Endpoint	2	f(2) = 2sin(2)+cos(2(2))=0.069+0.99=1.06

Global maximum of  f(x) = 2 sin(x) + cos(2x) on interval [0,2] is  3/2 at x = π/6

Global minimum of f(x) = 2 sin(x) + cos(2x) on interval [0,2] 1 at x = π/2 and x = 0.