Question

At what point in the interval [0,2π] does the function sin2x attain its maximum value?

Answer 1

Answer:

x = π/4 and 5π/4

Step-by-step explanation:

f(x) = Sin2x     x∈ [0,2π]

Find f'(x) first.

f'(x) = d/dx(Sin2x) = 2Cos2x.

Putting f'(x) = 0

2Cos2x = 0

Cos2x = 0

Cos2x = Cosπ/2

We know that if cosx = cosy, the general solution of x = 2nπ ± y

=> 2x = 2nπ + π/2

=> x = (2n+1)π/4

Putting n = 0 , x = π/4.

Putting n = 1, x = 3π/4

Putting n = 2, x = 5π/4.

Putting n = 3, x = 7π/4.

Putting n = 4, x = 9π/4

Putting n=5, x = 11π/4 > 2π.

Since  x∈ [0,2π]

The critical points are x = 0, π/4, 3π/4, 5π/4, 7π/4, 2π.

Value of f(x) at x = 0, π/4, 3π/4, 5π/4, 7π/4, 2π.

f(0) = 0

f(π/4) = 1.

f(3π/4) = - 1.

f(5π/4) = 1.

f(7π/4) = - 1.

f(2π) = 0.

Hence the maximum value of f(x) = 1 at x = π/4 and 5π/4