Problem 67 - Precalculus Mathematics for Calculus

Question

Find the maximum and minimum values of the function. $$y=\sin x+\sin 2 x$$

Step-by-Step Solution

Answer

The maximum value of the function is approximately 1.5 and the minimum is approximately -1.5.

Step 1: Rewrite the function using trigonometric identities

We start by rewriting the function using the double-angle identity for sine. The given function is \(y = \sin x + \sin 2x.\)Using the identity \(\sin 2x = 2\sin x \cos x,\) the function becomes:\[y = \sin x + 2\sin x \cos x = \sin x (1 + 2\cos x).\]This will help us simplify the analysis of the function.

Step 2: Find the first derivative

We need to find the critical points by setting the first derivative to zero. Use the product rule to differentiate:\(y(x) = \sin x (1 + 2\cos x)\).The derivative is:\[\frac{dy}{dx} = \cos x (1 + 2\cos x) - 2\sin^2 x.\]Simplify using \(\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1\), and we find:\[\frac{dy}{dx} = \cos x + 2\cos^2 x - 2\sin^2 x.\]\(= \cos x + 2(2\cos^2 x - 1 - \sin^2 x)\).

Step 3: Set the first derivative to zero

Set the derivative equal to zero to find the critical points:\[\cos x + 2(2\cos^2 x - 1 - \sin^2 x) = 0.\]Solving the equation involves substitution and simplification, where we use \(\sin^2 x = 1 - \cos^2 x\):\[\cos x + 4\cos^2 x - 2 = 0.\]This simplifies to solve for \(\cos x\).

Step 4: Solve for critical points

The equation \(\cos x + 4\cos^2 x - 2 = 0\) is quadratic in form. Substitute \(u = \cos x\) to get:\[4u^2 + u - 2 = 0.\]Solve this quadratic equation using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 1\), and \(c = -2\).

Step 5: Solve for \(u\) and substitute back to find \(x\)

Using the quadratic formula, \[u = \frac{-1 \pm \sqrt{1 + 32}}{8} = \frac{-1 \pm \sqrt{33}}{8}.\]These values for \(u\) give the values of \(\cos x\). Solve for \(x\) based on feasible trigonometric values within the range \([-1, 1]\).

Step 6: Evaluate function at critical points and endpoints

Substitute back the feasible \(x\) values into \(y = \sin x + \sin 2x \). Also check at endpoints (e.g., within one period, like \(0\) to \(2\pi\)) to ensure all extrema are found. Compare these values to determine maximum and minimum values of the function on this interval.

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools for simplifying expressions and solving equations involving trigonometric functions. They allow us to rewrite complex trigonometric expressions in more manageable forms. In this exercise, we use the double-angle identity:

• \( \sin 2x = 2\sin x \cos x \)

This identity helps to transform the original function \( y = \sin x + \sin 2x \) into the product \( y = \sin x (1 + 2\cos x) \). This form is more convenient for calculus operations such as differentiation. By understanding and applying trigonometric identities correctly, we can simplify not only the function's analysis but also ease the process of finding critical points and extrema.

Quadratic Equations

Quadratic equations often appear in trigonometric problems when looking for critical points, as seen in this exercise. We derived a quadratic equation in \( \cos x \) to find critical points:

• \( 4u^2 + u - 2 = 0 \), where \( u = \cos x \)

This quadratic equation can be solved using the quadratic formula:\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 4 \), \( b = 1 \), and \( c = -2 \). Quadratic equations like these help us find the specific trigonometric values that make the function's derivative zero, leading to potential critical points. Solving quadratic equations is a fundamental skill when analyzing the behavior of functions.

Critical Points

Critical points are where the derivative of a function equals zero or does not exist. They are crucial in determining where a function may reach a maximum or minimum. For the function \( y = \sin x + \sin 2x \), we found the derivative:\[ \frac{dy}{dx} = \cos x + 4\cos^2 x - 2 \]Setting this equal to zero, we solved for \( \cos x \) to find the critical points. These are the values where the function may have a peak or valley. Additionally, examining the derivative provides insights into the function's behavior, such as increasing or decreasing trends around these points. It's vital to check endpoints as well, since they might also reveal extreme values within a defined interval.

Function Analysis

Function analysis involves thoroughly examining a function to understand its behavior, including potential maximum and minimum values. By transforming and differentiating \( y = \sin x + \sin 2x \), we were able to identify critical points, necessary for locating extrema. However, applying function analysis also means evaluating the function at critical points and endpoints within a designated range (e.g., \( 0 \) to \( 2\pi \)).Key steps in function analysis encompass:

• Rewriting the function using identities for simplification.
• Finding and solving for critical points where the derivative equals zero.
• Evaluating the function at both critical points and endpoints.
• Comparing these values to determine relative and absolute extrema.

Comprehensive function analysis provides a full picture of the function's behavior over a specified interval, allowing us to deduce the maximum and minimum values effectively.