Problem 35 - Algebra and Trigonometry

Question

\(35-38=(a)\) Graph \(f\) and \(g\) in the given viewing rectangle and find the intersection points graphically, rounded to two decimal places. (b) Find the intersection points of \(f\) and \(g\) algebraically. Give exact answers. $$ \begin{array}{l}{f(x)=3 \cos x+1, g(x)=\cos x-1} \\ {[-2 \pi, 2 \pi] \text { by }[-2.5,4.5]}\end{array} $$

Step-by-Step Solution

Answer

Graphically: approx. \(x=-\pi,\pi\); Algebraically: exact \(x=\pi,-\pi\).

Step 1: Graph the Functions

To find the intersection points graphically, first, plot the functions \( f(x) = 3\cos x + 1 \) and \( g(x) = \cos x - 1 \) over the interval \([-2\pi, 2\pi]\). The viewing rectangle is from \([-2\pi, 2\pi]\) horizontally and \([-2.5, 4.5]\) vertically. Use graphing software or a calculator to create the plots, ensuring the y-axis ranges from -2.5 to 4.5.

Step 2: Identify Intersection Points Graphically

After graphing the functions, observe the points where the two graphs intersect. Zoom in to get a more accurate view if necessary. Record the x-coordinates of these intersection points, rounding them to two decimal places.

Step 3: Set Equations Equal for Algebraic Solution

To find the intersection points algebraically, set the equations \( f(x) = g(x) \). This gives \( 3\cos x + 1 = \cos x - 1 \). Simplify this equation to find the x-values where the two functions intersect.

Step 4: Solve the Equation

The equation \( 3\cos x + 1 = \cos x - 1 \) simplifies to \( 2\cos x = -2 \). Dividing both sides by 2 gives \( \cos x = -1 \). Find \( x \) values where the cosine of \( x \) equals -1 within the interval \([-2\pi, 2\pi]\).

Step 5: Identify Exact Solutions

Within the given interval, \( \cos x = -1 \) at \( x = \pi \) and \( x = -\pi \). These are the exact x-values where the functions intersect.

Key Concepts

Graphing Trigonometric Functions

To understand how to find intersection points of trigonometric functions, it's essential to start with graphing. Graphing provides a visual insight into the behavior of the functions over a specified interval. In this case, we are dealing with the functions \( f(x) = 3\cos x + 1 \) and \( g(x) = \cos x - 1 \). When graphing these functions over the interval \([-2\pi, 2\pi]\), we look at how the graphs oscillate based on the cosine function, which is periodic with a period of \(2\pi\).

• The function \( f(x) = 3\cos x + 1 \) is a scaled version of the standard cosine graph, meaning its amplitude is three times the usual one, and it is shifted upward by 1 unit.
• The function \( g(x) = \cos x - 1 \) has the standard amplitude for a cosine but shifted downward by 1 unit.

By plotting these functions, you can observe where their graphs intersect. These intersection points are crucial as they represent solutions to the equation \( f(x) = g(x) \) graphically. To ensure accuracy, zoom in if necessary and identify these points by their x-values.

Solving Trigonometric Equations

After graphing, solving trigonometric equations algebraically is the next step to precisely determine intersection points. Given the equations \( f(x) = 3\cos x + 1 \) and \( g(x) = \cos x - 1 \), we aim to find the x-values where the functions are equal. This leads to setting up the equation:\[ 3\cos x + 1 = \cos x - 1 \]By rearranging the terms, this simplifies to:\[ 2\cos x = -2 \]Through simplification, you identify values for which \( \cos x = -1 \). Solving trigonometric equations often involves combining algebraic manipulation with the unit circle understanding of trigonometric values.

Exact Solutions of Trigonometric Equations

Finding exact solutions to the equation \( \cos x = -1 \) within the interval \([-2\pi, 2\pi]\) involves recognizing key points on the unit circle where cosine, which is the x-coordinate, equals -1.

• On the unit circle, \( \cos x = -1 \) occurs at \( x = \pi \) and \( x = -\pi \). These points are directly across from the zero degrees or zero radians on the unit circle.
• Given the periodic nature of trigonometric functions, \( \cos x \) repeats its values every \(2\pi\). Within the specified interval, no repeated x-values exist beyond these two solutions.

Thus, the intersection points that solve the equation exactly are at \( x = \pi \) and \( x = -\pi \). Understanding these solutions helps you relate graphing results to algebraic calculations, ensuring a comprehensive grasp of trigonometric intersections.