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Find the intervals in which f(x) = sin 3x – cos 3x, 0 < x < π, is strictly increasing or strictly decreasing.

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Given function is

f(x) = sin 3x – cos 3x

f'(x) = sin 3x – cos 3x

For critical points of function f(x)

f'(x) = 0

Hence required possible intervals are

Hence, given function f(x) is strictly increasing in (0,\(\frac{\pi}{4}\)) ⋃ (\(\frac{7\pi}{12}\)\(\frac{11\pi}{12}\)

and strictly decreasing in  (\(\frac{\pi}{4}\),\(\frac{7\pi}{12}\)) ⋃ (\(\frac{11\pi}{12}\)\(\pi\)).

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HERE IS YOUR ANSWER

CORRECT QUESTION

Find the intervals in which the function f given by f(x) = 2x^2 − 3x is (a) strictly increasing (b) strictly decreasing

SOLUTION

Given, f(x) = 2x² - 3x
differentiate f(x) with respect to x,
f'(x) = 4x - 3 ------(1)

(a) when f(x) is strictly increasing function :
f'(x) > 0
from equation (1),
4x - 3 > 0 => x > 3/4
e.g., x ∈ (3/4, ∞ )
Therefore, the given function (f) is strictly increasing in interval x ∈ (3/4, ∞ ) .

(b) when f(x) is strictly decreasing function :
f'(x) < 0
from equation (1),
4x -3 < 0 => x < 3/4
e.g., x ∈ (-∞ , 3/4)
Therefore, the given function (f) is strictly decreasing in interval x ∈ (-∞ , 3/4)

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