Problem 54 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(41-48\) , consider the function on In Exercises \(49-54,\) (a) use a computer algebra system to differentiate the function, (b) sketch the graphs of \(f\) and \(f^{\prime}\) on the same set of coordinate axes over the given interval, (c) find the critical numbers of \(f\) in the open interval, and (d) find the interval(s) on which \(f^{\prime}\) is positive and the interval(s) on which it is negative. Compare the behavior of \(f\) and the sign of \(f^{\prime}\) $$ f(x)=2 \sin 3 x+4 \cos 3 x, \quad[0, \pi] $$

Step-by-Step Solution

Answer

The critical numbers of the function \( f(x) = 2 \sin 3x + 4 \cos 3x \) within the interval (0, \pi) are \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \). Also, \( f'(x) \) is positive when \( x \) is in the intervals (0, \( \frac{\pi}{6} \)) and (\( \frac{5\pi}{6} \), \pi) whereas it is negative when \( x \) is in the interval (\( \frac{\pi}{6} \), \( \frac{5\pi}{6} \)).

Step 1: Differentiate the function

Differentiate \( f(x) = 2 \sin 3x + 4 \cos 3x \) using the chain rule. The derivative of a sin or cos function is cos or sin respectively with the angle remaining the same and multiplied by the derivative of this angle. Thus, \(f'(x) = 6\cos3x - 12\sin3x \).

Step 2: Calculate the critical points

Set the derivative equation equal to zero and solve for x to get the critical numbers. So, \( 6\cos3x - 12\sin3x = 0 \)which simplifies as \( \tan3x = 0.5 \). By solving this equation, the critical numbers in the open interval (0, \pi) can be obtained. They are \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \)

Step 3: Determine Sign of the derivative function

Now plug in numbers within the intervals of the critical points into \( f'(x) \) to determine where \( f' \) is positive and negative. When \( x \) is between 0 and \( \frac{\pi}{6} \), \( \frac{5\pi}{6} \) and \( \pi \), \( f'(x) \) is positive. When \( x \) is between \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \), \( f'(x) \) is negative.

Step 4: Analyze and compare the behavior of f and sign of f'

The function \( f(x) \) increases when \( f'(x) \) is positive and decreases when \( f'(x) \) is negative. At the critical points \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \), \( f(x) \) changes its behaviour from increasing to decreasing and from decreasing to increasing respectively. This is in agreement with the First Derivative Test and the behavior of the derivative \( f'(x) \).

Key Concepts

Critical Numbers

When analyzing functions, critical numbers are crucial because they help us understand where a function might change its behavior, such as turning from increasing to decreasing or vice-versa. These numbers occur at points where the derivative is zero or undefined. For the given exercise, we dealt with the function \(f(x) = 2\sin3x + 4\cos3x\) over the interval \([0, \pi]\).
To find critical numbers, we first differentiate the function to get its derivative, \(f'(x)\), and then set \(f'(x) = 0\) because these are the potential points where changes in behavior can happen. After solving \(6\cos3x - 12\sin3x = 0\), we obtained \(\tan3x = 0.5\), resulting in critical numbers at \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\), within the interval (0, \pi). This process helps in identifying places where the function might have local maxima or minima.

Derivative of Trigonometric Functions

Differentiating trigonometric functions can at first seem tricky, but understanding the basic derivatives simplifies the process. For sine and cosine functions, the rules are straightforward:

• The derivative of \(\sin u\) is \(\cos u \cdot u'\),
• And the derivative of \(\cos u\) is \(-\sin u \cdot u'\).

When we applied these rules to \(f(x) = 2\sin3x + 4\cos3x\), we used the chain rule to consider the derivative of the inner function \(3x\), giving us \(f'(x) = 6\cos3x - 12\sin3x\).
This derivative combines the regular derivative process with the chain rule, making it vital for obtaining accurate results, especially when dealing with composite functions.

Chain Rule

The Chain Rule is a key tool for differentiating composite functions, allowing us to manage functions nested within others. For example, in \(f(x) = 2\sin3x + 4\cos3x\), we recognize that the trigonometric functions have an inner function, \(3x\).
To correctly differentiate this, the Chain Rule tells us to first differentiate the outer function while keeping the inner function unchanged, then multiply by the derivative of the inner function. Therefore:

• Differentiating \(\sin(3x)\) yields \(\cos(3x)\cdot 3\),
• And differentiating \(\cos(3x)\) gives us \(-\sin(3x)\cdot 3\).

By combining these results, we form the complete derivative \(f'(x) = 6\cos3x - 12\sin3x\). The Chain Rule simplifies handling such problems, especially when multiple function layers are involved.

Behavior of Functions

Understanding the behavior of a function is essential to graphing it and predicting its characteristics, such as increases or decreases, and potential turning points. The sign of a derivative like \(f'(x)\) is what helps us determine these behavior changes.
When \(f'(x) > 0\), the function \(f(x)\) is increasing. When \(f'(x) < 0\), it’s decreasing. For the given problem:

• Function \(f(x)\) increases between the intervals \((0, \frac{\pi}{6})\) and \((\frac{5\pi}{6}, \pi)\).
• It decreases between \((\frac{\pi}{6}, \frac{5\pi}{6})\).

The critical numbers, \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) are points where the function transitions from increasing to decreasing, observed in its first derivative \(f'(x)\).
These transition points align with the First Derivative Test, allowing us to correlate a function’s graphical behavior with its algebraic sign changes.