Fine the maximum and minimum value of f(x)=x+sin2x in the invervat 0,2π
Answers
Answer:
Minimum is 0 at x = 0.
Maximum is 2π at x = 2π.
Step-by-step explanation:
The extrema occur at critical points, so where the derivative is zero or undefined, or at an endpoint.
f(x) = x + sin 2x => f'(x) = 1 + 2 cos 2x
So
f'(x) = 0 => cos 2x = -1/2
=> 2x = 2π/3, 4π/3, 8π/3 or 10π/3 (since 2x can be in the interval [0,4π] )
=> x = π/3, 2π/3, 4π/3 or 5π/3.
Including the endpoints, the critical points are therefore:
0, π/3, 2π/3, 4π/3, 5π/3, 2π
and the extrema must occur at these places.
Checking the values at each:
f(0) = 0 + 0 = 0
f(π/3) = π/3 + sin(2π/3) = π/3 + √3/2 ≈ 1.913
f(2π/3) = 2π/3 + sin(4π/3) = 2π/3 - √3/2 ≈ 1.228
f(4π/3) = 4π/3 + sin(8π/3) = 4π/3 + √3/2 ≈ 5.055
f(5π/3) = 5π/3 + sin(10π/3) = 5π/3 - √3/2 ≈ 4.370
f(2π) = 2π + sin(4π) = 2π ≈ 6.283
Reading off from here:
- the minimum is f(0) = 0
- the maximum is f(2π) = 2π