Finding Extreme Values of Functions

To find the number of extreme values (local maxima or minima) of a function \(f(x)\) in an open interval, we typically follow these steps:

1. Find the first derivative of the function, \(f'(x)\).
2. Find the critical points by setting the first derivative equal to zero, i.e., solving \(f'(x) = 0\) for \(x\).
3. Determine which of these critical points lie within the specified domain (open interval).
4. For functions that are differentiable over an open interval, the critical points within that interval are the locations of potential extreme values. The number of distinct critical points within the open interval usually corresponds to the number of extreme values.

Analyzing the Given Function sin4x + 2x

The given function is \(f(x) = \sin(4x) + 2x\). We are interested in the number of extreme values in the domain \(0 < x < \frac{\pi}{2}\).

First, let's find the derivative of \(f(x)\) with respect to \(x\):

\[ f'(x) = \frac{d}{dx}(\sin(4x) + 2x) \]

Using the chain rule for \(\sin(4x)\) and the power rule for \(2x\):

\[ f'(x) = \cos(4x) \cdot \frac{d}{dx}(4x) + 2 \]\[ f'(x) = \cos(4x) \cdot 4 + 2 \]\[ f'(x) = 4\cos(4x) + 2 \]

Finding Critical Points for sin4x + 2x

Next, we find the critical points by setting \(f'(x) = 0\):

\[ 4\cos(4x) + 2 = 0 \]

Now, we solve this equation for \(x\):

\[ 4\cos(4x) = -2 \]\[ \cos(4x) = \frac{-2}{4} \]\[ \cos(4x) = -\frac{1}{2} \]

Solving for x in the Domain \(0 < x < \frac{\pi}{2}\)

We need to find the values of \(4x\) such that \(\cos(4x) = -\frac{1}{2}\) within the relevant interval for \(4x\). The given domain for \(x\) is \(0 < x < \frac{\pi}{2}\).

Multiplying the inequality by 4, we get the domain for \(4x\):

\[ 4 \cdot 0 < 4x < 4 \cdot \frac{\pi}{2} \]\[ 0 < 4x < 2\pi \]

We are looking for angles \(\theta\) such that \(\cos(\theta) = -\frac{1}{2}\) in the interval \( (0, 2\pi) \). The cosine function is negative in the second and third quadrants. The basic angle for which \(\cos(\theta) = \frac{1}{2}\) is \(\frac{\pi}{3}\).

• In the second quadrant, the angle is \(\pi - \frac{\pi}{3} = \frac{3\pi - \pi}{3} = \frac{2\pi}{3}\).
• In the third quadrant, the angle is \(\pi + \frac{\pi}{3} = \frac{3\pi + \pi}{3} = \frac{4\pi}{3}\).

These are the values for \(4x\). So, we have:

\[ 4x = \frac{2\pi}{3} \quad \text{or} \quad 4x = \frac{4\pi}{3} \]

Now, we solve for \(x\):

• For the first value: \(x = \frac{1}{4} \cdot \frac{2\pi}{3} = \frac{2\pi}{12} = \frac{\pi}{6}\).
• For the second value: \(x = \frac{1}{4} \cdot \frac{4\pi}{3} = \frac{4\pi}{12} = \frac{\pi}{3}\).

Checking if Critical Points are in the Domain \(0 < x < \frac{\pi}{2}\)

We need to verify if the calculated values of \(x\) are within the specified domain \(0 < x < \frac{\pi}{2}\).

• For \(x = \frac{\pi}{6}\): Is \(0 < \frac{\pi}{6} < \frac{\pi}{2}\)? Yes, because \(\frac{\pi}{2} = \frac{3\pi}{6}\), and \(0 < \frac{\pi}{6} < \frac{3\pi}{6}\) is true.
• For \(x = \frac{\pi}{3}\): Is \(0 < \frac{\pi}{3} < \frac{\pi}{2}\)? Yes, because \(\frac{\pi}{3} = \frac{2\pi}{6}\), and \(0 < \frac{2\pi}{6} < \frac{3\pi}{6}\) is true.

Both critical points, \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{3}\), lie within the domain \(0 < x < \frac{\pi}{2}\).

Conclusion on Number of Extreme Values

Since we found exactly two critical points within the open interval \(0 < x < \frac{\pi}{2}\), and the function is differentiable in this interval, each of these critical points corresponds to an extreme value (either a local maximum or a local minimum). Therefore, the function \(f(x) = \sin(4x) + 2x\) has 2 extreme values in the interval \(0 < x < \frac{\pi}{2}\).

Revision Table: Extreme Values and Critical Points

Additional Information: Trigonometric Equations in Calculus

Solving trigonometric equations is a common task when finding critical points of functions involving trigonometric terms. Remember the general solutions for trigonometric equations like \(\cos(\theta) = c\):

• For \(\cos(\theta) = c\), where \(-1 \le c \le 1\), if \(\alpha\) is the principal value such that \(\cos(\alpha) = c\), then the general solution is \(\theta = 2n\pi \pm \alpha\), where \(n\) is an integer.

In our case, \(\cos(4x) = -\frac{1}{2}\). The principal value for \(\cos(\alpha) = -\frac{1}{2}\) is \(\alpha = \frac{2\pi}{3}\) (in the interval \( [0, \pi] \)). The general solution for \(4x\) is:

\[ 4x = 2n\pi \pm \frac{2\pi}{3} \]\[ x = \frac{2n\pi}{4} \pm \frac{2\pi}{12} \]\[ x = \frac{n\pi}{2} \pm \frac{\pi}{6} \]

We need to find integer values of \(n\) such that \(0 < x < \frac{\pi}{2}\). Let's test values of \(n\):

• If \(n = 0\): \(x = 0 \pm \frac{\pi}{6} \implies x = \frac{\pi}{6}\) or \(x = -\frac{\pi}{6}\). Only \(x = \frac{\pi}{6}\) is positive. Check \(0 < \frac{\pi}{6} < \frac{\pi}{2}\). This is true.
• If \(n = 1\): \(x = \frac{\pi}{2} \pm \frac{\pi}{6}\). \(x = \frac{\pi}{2} + \frac{\pi}{6} = \frac{3\pi + \pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}\) (Too large, \(\frac{2\pi}{3} > \frac{\pi}{2}\)). \(x = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}\). Check \(0 < \frac{\pi}{3} < \frac{\pi}{2}\). This is true.
• If \(n < 0\) or \(n > 1\), the resulting \(x\) values will fall outside the range \( (0, \frac{\pi}{2}) \). For example, for \(n=-1\), \(x = -\frac{\pi}{2} \pm \frac{\pi}{6}\) gives negative values. For \(n=2\), \(x = \pi \pm \frac{\pi}{6}\) gives values greater than \(\frac{\pi}{2}\).

This confirms that the only solutions in the interval \( (0, \frac{\pi}{2}) \) are \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{3}\). There are 2 such values.