Show that f(x) = cos x is a decreasing function on (0, π), increasing

1.	Show that f(x) = cos x is a decreasing function on (0, π), increasing in (–π, 0) and neither increasing nor decreasing in (–π, π).
Answer» Given as f(x) = cos x f'(x) = (d/dx)(cos x) ⇒ f’(x) = –sin x On taking different region from 0 to 2π Suppose x ∈ (0, π). ⇒ Sin(x) > 0 ⇒ –sin x < 0 ⇒ f’(x) < 0 Hence, f(x) is decreasing in (0, π) Suppose x ∈ (–π, o). ⇒ Sin (x) < 0 ⇒ –sin x > 0 ⇒ f’(x) > 0 Hence f(x) is increasing in (–π, 0). So, from above condition we find that ⇒ f (x) is decreasing in (0, π) and increasing in (–π, 0). Thus, condition for f(x) neither increasing nor decreasing in (–π, π)

Show that f(x) = cos x is a decreasing function on (0, π), increasing in (–π, 0) and neither increasing nor decreasing in (–π, π).

Answer»

Given as f(x) = cos x

f'(x) = (d/dx)(cos x)

⇒ f’(x) = –sin x

On taking different region from 0 to 2π

Suppose x ∈ (0, π).

⇒ Sin(x) > 0

⇒ –sin x < 0

⇒ f’(x) < 0

Hence, f(x) is decreasing in (0, π)

Suppose x ∈ (–π, o).

⇒ Sin (x) < 0

⇒ –sin x > 0

⇒ f’(x) > 0

Hence f(x) is increasing in (–π, 0).

So, from above condition we find that

⇒ f (x) is decreasing in (0, π) and increasing in (–π, 0).

Thus, condition for f(x) neither increasing nor decreasing in (–π, π)