Finding the Longest Increasing Interval of a Trigonometric Function

To find the intervals where a function is increasing, we need to analyze its derivative. A function \(f(x)\) is increasing on an interval if its derivative \(f'(x) > 0\) for all \(x\) in that interval.

Step 1: Find the Derivative of the Function

The given function is \(f(x) = 3\sin x - 4\sin^3 x\). This expression is a known trigonometric identity for \(\sin(3x)\). So, we can rewrite the function as \(f(x) = \sin(3x)\).

Now, let's find the derivative \(f'(x)\) using the chain rule:

\[f'(x) = \frac{d}{dx}(\sin(3x))\]

Using the chain rule, if \(y = \sin(u)\) where \(u = 3x\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

• \(\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u = \cos(3x)\)
• \(\frac{du}{dx} = \frac{d}{dx}(3x) = 3\)

So, the derivative is:

\[f'(x) = \cos(3x) \cdot 3 = 3\cos(3x)\]

Step 2: Determine Where the Function is Increasing

The function \(f(x)\) is increasing when \(f'(x) > 0\).

\[3\cos(3x) > 0\]

Dividing by 3 (a positive constant), the inequality remains the same:

\[\cos(3x) > 0\]

Step 3: Find the Intervals for \(3x\) where \(\cos(3x) > 0\)

The cosine function is positive in the first and fourth quadrants. The general solution for \(\cos(\theta) > 0\) is:

\[-\frac{\pi}{2} + 2n\pi < \theta < \frac{\pi}{2} + 2n\pi\]

where \(n\) is an integer (\(n \in \mathbb{Z}\)).

In our case, \(\theta = 3x\). So, we have:

\[-\frac{\pi}{2} + 2n\pi < 3x < \frac{\pi}{2} + 2n\pi\]

Step 4: Solve for \(x\) to Find the Intervals of Increase

Divide the inequality by 3:

\[\frac{1}{3}\left(-\frac{\pi}{2} + 2n\pi\right) < \frac{1}{3}(3x) < \frac{1}{3}\left(\frac{\pi}{2} + 2n\pi\right)\]

\[-\frac{\pi}{6} + \frac{2n\pi}{3} < x < \frac{\pi}{6} + \frac{2n\pi}{3}\]

These are the intervals where the function \(f(x)\) is increasing. Let's list some of these intervals by substituting different integer values for \(n\):

• For \(n=0\): \(-\frac{\pi}{6} < x < \frac{\pi}{6}\), Interval: \(\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)\)
• For \(n=1\): \(-\frac{\pi}{6} + \frac{2\pi}{3} < x < \frac{\pi}{6} + \frac{2\pi}{3}\) \(-\frac{\pi}{6} + \frac{4\pi}{6} < x < \frac{\pi}{6} + \frac{4\pi}{6}\) \(\frac{3\pi}{6} < x < \frac{5\pi}{6}\), Interval: \(\left(\frac{\pi}{2}, \frac{5\pi}{6}\right)\)
• For \(n=-1\): \(-\frac{\pi}{6} - \frac{2\pi}{3} < x < \frac{\pi}{6} - \frac{2\pi}{3}\) \(-\frac{\pi}{6} - \frac{4\pi}{6} < x < \frac{\pi}{6} - \frac{4\pi}{6}\) \(-\frac{5\pi}{6} < x < -\frac{3\pi}{6}\), Interval: \(\left(-\frac{5\pi}{6}, -\frac{\pi}{2}\right)\)

And so on.

Step 5: Calculate the Length of Each Interval

The length of a general interval \(\left(-\frac{\pi}{6} + \frac{2n\pi}{3}, \frac{\pi}{6} + \frac{2n\pi}{3}\right)\) is given by the upper bound minus the lower bound:

\[\text{Length} = \left(\frac{\pi}{6} + \frac{2n\pi}{3}\right) - \left(-\frac{\pi}{6} + \frac{2n\pi}{3}\right)\]

\[\text{Length} = \frac{\pi}{6} + \frac{2n\pi}{3} + \frac{\pi}{6} - \frac{2n\pi}{3}\]

\[\text{Length} = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}\]

The length of each interval of increase is \(\frac{\pi}{3}\).

Step 6: Determine the Length of the Longest Interval

Since all the intervals of increase have the same length, \(\frac{\pi}{3}\), the length of the longest interval in which the function is increasing is \(\frac{\pi}{3}\).

Conclusion

The length of the longest interval in which the function \(f(x) = 3\sin x - 4\sin^3 x\) is increasing is \(\frac{\pi}{3}\).

Comparing this result with the given options:

• Option 1: \(\frac{{\rm{\pi }}}{3}\)
• Option 2: \(\frac{{\rm{\pi }}}{2}\)
• Option 3: \(\frac{{3{\rm{\pi }}}}{2}\)
• Option 4: \(\pi\)

The calculated length matches Option 1.

Revision Table: Function Increasing Interval

Additional Information: Analyzing Function Behavior

Understanding where a function is increasing or decreasing is crucial in calculus. This analysis helps in sketching the graph of the function and finding local maxima and minima.

• Monotonicity: A function is called monotonic on an interval if it is either entirely increasing or entirely decreasing on that interval.
• Critical Points: Points where \(f'(x) = 0\) or \(f'(x)\) is undefined. These points often mark transitions between increasing and decreasing intervals.
• First Derivative Test: This test uses the sign of the first derivative to determine whether a critical point corresponds to a local maximum, local minimum, or neither. If \(f'(x)\) changes from positive to negative at a critical point, there's a local maximum. If it changes from negative to positive, there's a local minimum. If the sign doesn't change, it's neither.
• Periodicity: Trigonometric functions are periodic. The function \(f(x) = \sin(3x)\) has a period of \(\frac{2\pi}{3}\). This means its behavior, including increasing and decreasing intervals, repeats every \(\frac{2\pi}{3}\) units on the x-axis. The intervals of increase we found also repeat with a pattern corresponding to this period.

The problem leveraged the identity \(3\sin x - 4\sin^3 x = \sin(3x)\). Recognizing such identities can significantly simplify the differentiation process. If this identity was not recognized, we would have used the derivative of \(\sin^3 x\) which is \(3\sin^2 x \cos x\) and proceeded as initially shown in the thought process, leading to \(f'(x) = 3\cos x - 12\sin^2 x \cos x = 3\cos x (1 - 4\sin^2 x)\). Using the identity \(1 - 2\sin^2 x = \cos(2x)\), we can write \(1 - 4\sin^2 x = 1 - 2(2\sin^2 x) = 1 - 2(1 - \cos(2x)) = 1 - 2 + 2\cos(2x) = 2\cos(2x) - 1\). This would give \(f'(x) = 3\cos x (2\cos(2x) - 1)\). Analyzing the sign of this expression would be more complex than analyzing \(3\cos(3x)\). Using the trigonometric identity simplified the problem significantly.