Question

Find the maximum and minimum values of f(x)=sin⁴x+cos⁴x. x ∈ [0,π/2]

Answer 1

we have to find maximum and minimum values of f(x) = sin⁴x + cos⁴x where x ∈ [0, π/2]

differentiating f(x) with respect to x,

f'(x) = 4sin³x.cosx + 4cos³x (-sinx)

= 4sinx.cosx ( sin²x - cos²x)

= -4sinx.cosx(cos²x - sin²x)

we know,

2sinθ.cosθ = sin2θ

so, 2sinx.cosx = sin2x

and cos²θ - sin²θ = cos2θ

so, cos²x - sin²x = cos2x

then, f'(x) = -2sin2x.cos2x = -sin4x

f'(x) = -sin4x = 0

⇒4x = nπ

⇒x = nπ/4

but in [0, π/2 ] , x = π/4

now we have. three points 0, π/4 and π/2 to check maxima and minima.

differentiate f'(x) with respect to x,

f"(x) = -4 × 4cos4x = -16cos4x

at x = 0, f"(0) = -16cos4(0) = -16 < 0 and f'(0) = -sin4(0) = 0

so, f(x) will be maximum at x = 0

and maximum value of f(x) =sin⁴(0) + cos⁴(0) = 0 + 1 = 1

[ note : function will be maximum also at x = π/2, and maximum value is not other than 1. you can check on it ]

at x = π/4, f"(π/4) = -16cos4(π/4) = 16> 0

and f'(π/4) = -4sin4(π/4) = 0

so, f(x) will be minimum at x = π/4.

and minimum value of f(x) = sin⁴(π/4) + cos⁴(π/4) = 1/4 + 1/4 = 1/2