Question

Find the intervals of monotonicities and hence find the local extremum for the following functions:

f(x) = sin x cos x + 5, x ∈ (0, 2π)

Answer 1

f'(x) = cos 2x

f'(x) = 0 ⇒ cos 2x = 0

Stationary points

x = `pi/4, (3pi)/4, (5pi)/4, pi/4 ∈x = (0, 2π)`

The intervals are `(0, pi/4), (pi/4, (3pi)/4)((3pi)/4, (5pi)/4)((5pi)/4, (7pi)/4)` and `((7pi)/4, 2pi)`

In the interval `(0, pi/4)`, f'(x) > 0 ⇒ f(x) is strictly increasing.

In the interval `(pi/4, (3pi)/4)`, f'(x) < 0 ⇒ f(x) is strictly decreasing.

In the interval `((3pi)/4, (5pi)/4)`, f'(x) > 0 ⇒ f(x) is strictly increasing.

In the interval `((5pi)/4, (7pi)/4)`, f'(x) < 0 ⇒ f(x) is strictly decreasing.

In the interval `((7pi)/4, 2pi)`, f'(x) > 0 ⇒ f(x) is strictly increasing.

f'(x) changes its sign from positive to negative when passing through x = `pi/4` and x = `(5pi)/4`

∴ f(x) attains local maximum at x = `pi/4` and `(5pi)/4`

∴ Local maximum `"f"(pi/4)`

= `sin(pi/4) cos (pi/4) + 5`

= `1/sqrt(2) 1/sqrt(2) + 5`

= `1/2 + 5`

= `11/2`

And `"f"((5pi)/4) = sin((5pi)/4) cos  ((5pi)/4) + 5`

= `(- sqrt(2)/2)(- sqrt(2)/2) + 5`

= `11/2`

f'(x) changes its sign from negative to positive when passing through x = `(3pi)/4` and x = `(7pi)/4`

∴ f(x) attains local maximum at x = `(3pi)/4` and x = `(5pi)/4`

∴ Local minimum `"f"((3pi)/4)`

= `sin((3pi)/4) cos ((3pi)/4) + 5`

= `(sqrt(2)/2)(- sqrt(2)/2) + 5`

= `- 1/2 + 5`

= `9/2`

And `"f"((7pi)/4) = (- sqrt(2)/2)(sqrt(2)/2) + 5`

= `- 1/2 + 5`

= `9/2`