Question

Find maximum and minimum values of f(x)=x-2sinx x ∈ [0, 2π]

Answer 1

function, f(x) = x - 2sinx

differentiate f(x) with respect to x,

f'(x) = 1 - 2cosx

now, f'(x) = 1 - 2cosx = 0

⇒1 - 2cosx = 0

⇒cosx = 1/2 = cos(π/6)

⇒x = π/6, 11π/6

[ as you know, cosine function is positive in first and 4th quadrants. so, π/6 and 11π/6 is the solution of cosx = 1/2 in [0, 2π] ]

now again differentiate with respect to x,

f"(x) = 2sinx

at x = π/6 , f"(π/6) = 2sin(π/6) = 1 > 0

hence, f(x) will be minimum at x = π/6

so, minimum value of f(x) = f(π/6) = π/6 - 2sin(π/6) = π/6 - 1

at x = 11π/6, f"(11π/6) = 2sin(11π/6) = -1 < 0

hence, f(x) will be maximum at x = 11π/6

so, maximum value of f(x) = f(11π/6) = 11π/6 - 2sin(11π/6) = 11π/6 + 1