Question

Solve the following: 

Find the maximum and minimum values of the function f(x) = cos²x + sinx.

Answer 1

f(x) = cos²x + sinx

∴ f'(x) = `d/dx(cos^2x + sinx)`

= `2cosx.d/dx(cosx) + cosx`

= 2 cosx (–sin x) + cosx

= –sin 2x + cos x

and f"(x) = `d/dx(- sin 2x) + cosx`

= `-cos2x.d/dx(2x) - sinx`

= –cos 2x × 2 – sinx

= – 2 cos 2x – sinx

For extreme values of f(x), f'(x) = 0

∴ –sin 2x + cos x = 0

∴ –2 sin x cos x + cos x = 0

∴ cos x (–2sin x + 1) = 0

∴ cos x = 0 or –2 sin x + 1 = 0

∴ cos x = `cos  pi/(2) or sin x = (1)/(2) = sin  pi/(6)`

∴ x =  `pi/(2)` or x =  `pi/(6)`

(i) `f"(pi/2) = 2cospi - sin  pi/(2)`

= –2(–1) – 1 = 1 > 0

∴ By the second derivative test, f is minimum at x = `pi/(2)` and minimum value of f at x = `pi/(2)`

= `f(pi/2) = cos^2  pi/(2) + sin  pi/(2) = 0 + 1 = 1`

(ii) `f^"(pi/6) = - 2 cos  pi/(3) - sin  pi/(6)`

= `-2(1/2) - (1)/(2)`

= `-(3)/(2) < 0`

∴ By the second derivative test, f is maximum at x = `pi/(6)` and maxmuum value of f at x = `pi/(6)` is

= `f(pi/6) = cos^2  pi/(6) + sin  pi/(6)`

= `(sqrt(3)/2)^2 + (1)/(2) = (5)/(4)`

Hence, the maximum and minimum values of the 5 function f(x) are `(5)/(4)` and 1 respectively.