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Question
Solve the following:
Find the maximum and minimum values of the function f(x) = cos2x + sinx.
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Solution
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.
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