Question

Find the interval in which f(x)=cos 3x iw strictly increasing or decreasing

Answer 1

Answer:

increasing x = ( π/3 to 2π/3  + 2πn/3)

Decreasing x = ( 0 to π/3  + 2πn/3)

Step-by-step explanation:

Find the interval in which f(x)=cos 3x iw strictly increasing or decreasing

f(x)=cos 3x

f(x) = Cos(0°) = 1

=> 3x = 0°

=> x = 0°

f(x) = Cos(90°) = 0

=> 3x = 90°

=> x = 30° = π/6

f(x) = Cos(180°) = -1

=> 3x = 180°

=> for x = 60° = π/3

for x from 0 to 60°

f(x) is from 1 to -1

f(x)=cos 3x is strictly decreasing for x from o to 60°  ( 0 to π/3)

Cos(3x) = Cos(3x + 2πn)

x = ( 0 to π/3  + 2πn/3)

Similarly

f(x) = Cos(360°) = 1

=> 3x = 360°

=> for x = 120° = π/3

so for x from 60°  to 120° (π/3 to 2π/3) f(x) is strictly increasing

x = ( π/3 to 2π/3  + 2πn/3)

Answer 2

Answer:

Concept:

First derivative test:

If f '(x) > 0 for all x in I, then f(x) is strictly increasing on I

If f '(x) < 0 for all x in I, then f(x) is strictly decreasing on I

f(x)= cos3x

f '(x)= - 3 sin3x

f '(x) = 0

⇒ -3 sin3x=0

⇒ sin3x =0

$3x = n\pi, n\:is\:an\:integer\\\\x=\frac{n\pi}{3}$

The critical numbers are

$0, \:\frac{\pi}{3}, \:\frac{2\pi}{3},\: \pi, \:\frac{4\pi}{3},\: \frac{5\pi}{3}, \:2\pi$

$when \:0<x<\frac{\pi}{3}, \:f '(x) < 0$

f(x) is strictly decreasing

$when \:\frac{\pi}{3}<x<\frac{2\pi}{3}, \:f '(x) > 0$

f(x) is strictly increasing

$when \:\frac{2\pi}{3}<x<\pi, \:f '(x) < 0$

f(x) is strictly decreasing

$when \:\pi<x<\frac{4\pi}{3}, \:f '(x) > 0$

f(x) is strictly increasing

$when \:\frac{4\pi}{3}<x<\frac{5\pi}{3}, \:f '(x) < 0$

f(x) is strictly decreasing

$when \:\frac{5\pi}{3}<x<2\pi, \:f '(x) > 0$

f(x) is strictly increasing