Problem 41 - Calculus Early Transcendentals

Question

a. Find the critical points of \(f\) on the given interval. b. Determine the absolute extreme values of \(f\) on the given interval when they exist. c. Use a graphing utility to confirm your conclusions. $$f(x)=\sin 3 x \text { on }[-\pi / 4, \pi / 3]$$

Step-by-Step Solution

Answer

2. What is the critical point of the function \(f(x) = \sin(3x)\) in the interval \([-\pi/4, \pi/3]\)? 3. What are the absolute minimum and maximum values of the function \(f(x) = \sin(3x)\) in the interval \([-\pi/4, \pi/3]\)? 4. Why is it helpful to confirm the conclusions with a graphing utility?

Step 1: Find the first derivative of f(x)

Calculate the first derivative of \(f(x) = \sin(3x)\) using the chain rule, which states that \((f(g(x)))' = f'(g(x))g'(x)\). Let \(u = 3x\), then \(f(u) = \sin(u)\). Thus, we have: $$f'(u) = \cos(u)\ \text{and}\ u'(x) = 3$$ Using the chain rule, we get: $$f'(x) = f'(u)u'(x) = \cos(3x) \cdot 3 = 3\cos(3x)$$.

Step 2: Find the critical points

Now, set the first derivative to zero to find the critical points: $$3\cos(3x) = 0$$ Divide by 3: $$\cos(3x) = 0$$ Now, solve for x within the interval \([-\pi/4, \pi/3]\): $$3x = \frac{\pi}{2} + k\pi$$ where \(k\) is an integer. Solve for x: $$x = \frac{\pi}{6} + \frac{k\pi}{3}$$ Now, find the integer values of k that give x values within the interval: $$-\frac{\pi}{4} \leq \frac{\pi}{6} + \frac{k\pi}{3} \leq \frac{\pi}{3}$$ $$-\frac{\pi}{12}-\frac{k\pi}{3} \leq 0 \leq -\frac{\pi}{12}+\frac{k\pi}{3}$$ For k = 0: \(x = \pi/6\) is within the interval and is a critical point. For k = -1, the expression is always negative , while for k = 1, the expression is always positive, so there are no further critical points. Thus, the only critical point is \(x = \pi/6\).

Step 3: Determine absolute extreme values

Evaluate the function at the critical point and the interval's endpoints: $$f(-\pi/4) = \sin\left(3\cdot\left(-\frac{\pi}{4}\right)\right) = \sin\left(-\frac{3\pi}{4}\right)$$ $$f(\pi/6) = \sin\left(3\cdot\left(\frac{\pi}{6}\right)\right) = \sin(\pi) = 0$$ $$f(\pi/3) = \sin\left(3\cdot\left(\frac{\pi}{3}\right)\right) = \sin(\pi) = 0$$ The absolute minimum is \(f(-\pi/4) = \sin(-3\pi/4) \approx -0.7071\), and the absolute maximum is \(f(\pi/6) = f(\pi/3) = 0\).

Step 4: Confirm conclusions with a graphing utility

Using a graphing utility (e.g., Desmos, GeoGebra, or a graphing calculator), plot the function \(f(x) = \sin(3x)\) on the interval \([-\pi/4, \pi/3]\) . The graph should show a local minimum at \(x=-\pi/4\) and a local maximum at \(x = \pi/6\). Additionally, the graph should confirm that the absolute minimum and maximum are as we found in step 3. By visually inspecting the graph, you'll confirm that the conclusions are correct.

Key Concepts

First Derivative Test

To determine the behavior of a function and identify points of interest, such as local maxima and minima, the First Derivative Test is a vital tool. The first derivative of a function indicates whether the original function is increasing or decreasing at a given point. Here's how it works:

• If the first derivative changes from positive to negative, the function has a local maximum at that point.
• Conversely, if it changes from negative to positive, there is a local minimum.
• If there’s no change in sign, the test is inconclusive or the point could be an inflection point.

Applying this to our example function, \( f'(x) = 3 \cos(3x) \), we set the derivative to zero to identify critical points, solving \( 3 \cos(3x) = 0 \). This helps locate potential maxima or minima within the specified interval, confirming the presence of these features through further testing or graphical analysis.

Extreme Value Theorem

The Extreme Value Theorem is essential when working with continuous functions over a closed interval. It ensures that such functions will always reach a minimum and a maximum value. In simpler terms, if you graph a continuous function over a closed interval, you’ll find peaks and valleys at specific points. For the function \(f(x) = \sin(3x)\)on the interval \([-\pi/4, \pi/3]\), we calculate the function's value at:

• The endpoints \(-\pi/4\) and \(\pi/3\)
• The critical point we find \(\pi/6\)

This allows us to verify where the maximum and minimum values occur. By checking the end and critical points, we confirm using calculations in the original solution, that \(f(\pi/6) = 0\),\(f(\pi/3) = 0\), and \(f(-\pi/4) \approx -0.7071\) are indeed the extreme values.

Chain Rule

The Chain Rule is a fundamental differentiation technique, especially crucial when dealing with composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function itself. For instance, when finding the derivative of \(f(x) = \sin(3x)\),you set \(u = 3x\),making the function \(\sin(u)\). The derivative is then:

• The derivative of \(\sin(u)\) is \(\cos(u)\)
• The derivative of \(u = 3x\) is \(3\)

So, using the Chain Rule, the derivative becomes \(3 \cos(3x)\), which is used to find critical points.

Graphical Analysis of Functions

Visualization can often simplify complex calculus concepts. Graphical analysis allows students to visually confirm the existence and location of critical points, maxima, and minima of functions. For the function \(f(x) = \sin(3x)\) on the interval \([-\pi/4, \pi/3]\), plotting the function provides a clear picture of how it behaves. This helps confirm:

• A local minimum at \(x = -\pi/4\)
• A local maximum at \(x = \pi/6\)

By graphing, students can easily visually confirm that these are the points of increased or decreased function value, validating the calculus-based findings. Tools like Desmos or GeoGebra are particularly helpful for graphing complex functions with ease, showing the whole behavior of the function across the interval.