A calculus problem by Ayushman Chahar

Calculus Level 3

The length of a longest interval in which the function f ( x ) = 3 sin ⁡ x − 4 sin ⁡ 3 x f(x)=3\sin x-4\sin^{3}x is increasing, is

π / 3 \pi/3 3 π / 2 3\pi/2 π / 2 \pi/2 π \pi

Ayushman Chahar by Ayushman Chahar , 21, India

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1 solution

Tom Engelsman
Jun 14, 2017

The above function equals f ( x ) = s i n ( 3 x ) f(x) = sin(3x) and satisfies:

− 1 ≤ s i n ( 3 x ) ≤ 1 ⇒ − π 2 ≤ 3 x ≤ π 2 ⇒ − π 6 ≤ x ≤ π 6 . -1 \le sin(3x) \le 1 \Rightarrow -\frac{\pi}{2} \le 3x \le \frac{\pi}{2} \Rightarrow -\frac{\pi}{6} \le x \le \frac{\pi}{6}.

So the length of the longest interval where f ( x ) f(x) increases equals π 6 − ( − π 6 ) = π 3 . \frac{\pi}{6} - (-\frac{\pi}{6}) = \boxed{\frac{\pi}{3}}.

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