Problem 36 - Precalculus Mathematics for Calculus

Question

(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. $$\begin{aligned} &f(x)=\sin 2 x+1, g(x)=2 \sin 2 x+1\\\ &[-2 \pi, 2 \pi] \text { by }[-1.5,3.5] \end{aligned}$$

Step-by-Step Solution

Answer

Intersection points are approximately at \(-\frac{\pi}{2}, \frac{\pi}{2}\). Algebraically, these points are exact.

Step 1: Graph the Functions

First, plot the functions \( f(x) = \sin(2x) + 1 \) and \( g(x) = 2\sin(2x) + 1 \) on the coordinate plane with the domain \([-2\pi, 2\pi]\) and the range \([-1.5, 3.5]\). Notice that these functions are vertically stretched and shifted sine waves. Use a graphing tool to visualize them.

Step 2: Find Intersection Points Graphically

Using the graph plotted in Step 1, identify where the curves \( f(x) \) and \( g(x) \) intersect. For this particular exercise, graphically approximate the points where the curves meet. The approximate intersection points are at \( x = -\frac{\pi}{2}, \frac{\pi}{2} \) for this viewing window.

Step 3: Set Up the Equation for Algebraic Solution

To find the intersection points algebraically, set \( f(x) = g(x) \), i.e., \( \sin(2x) + 1 = 2\sin(2x) + 1 \). Simplify this equation to find where the functions intersect.

Step 4: Simplify the Equation

Simplify the equation \( \sin(2x) + 1 = 2\sin(2x) + 1 \) to \( 0 = \sin(2x) \). This implies that \( \sin(2x) = 0 \).

Step 5: Solve for x Algebraically

Solve the equation \( \sin(2x) = 0 \) to find the exact points where the functions intersect. The solutions for \( 2x = n\pi \) where \( n \) is an integer are \( x = \frac{n\pi}{2} \). Considering the domain \([-2\pi, 2\pi]\), the values for \( x \) are \( -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \).

Step 6: Verify Solutions

Verify the algebraic solutions by checking if they fall within the given domain \([-2\pi, 2\pi]\). The solutions \( x = -\frac{\pi}{2}, \frac{\pi}{2} \) are valid within this range and match the graphical intersections.

Key Concepts

Graphical Analysis

When dealing with trigonometric functions, visualizing their behavior on a graph can provide insightful understanding of their patterns and intersections. Start by plotting the functions \( f(x) = \sin(2x) + 1 \) and \( g(x) = 2\sin(2x) + 1 \) within the domain \([-2\pi, 2\pi]\) and the range \([-1.5, 3.5]\). This graphical representation transforms the abstract algebraic functions into a tangible visual image.

• Notice how both functions are sine waves, but with different amplitudes. Function \( f(x) \) has an amplitude slightly modified by the addition of 1.
• Function \( g(x) \) as a sine wave also shifts vertically by 1 and stretches vertically even more compared to \( f(x) \).

The visual intersection points are crucial as they indicate where \( f(x) \) and \( g(x) \) share the same value for a certain \( x \) in the viewing window. Approximating visually, we see intersections at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \). It's essential to be precise with these graphical estimates, often confirming them through algebraic methods or using graphing tools that allow you to trace points.

Algebraic Solution

Finding intersections algebraically involves setting the two functions equal and solving for \( x \). Start with the equation derived from their equality:

\[ f(x) = g(x) \Rightarrow \sin(2x) + 1 = 2\sin(2x) + 1 \]

By simplifying, we reach \( 0 = \sin(2x) \), so we need to solve for \( x \) when \( \sin(2x) = 0 \). The equation \( \sin(kx) = 0 \) is satisfied when \( kx = n\pi \) for some integer \( n \). Hence,

• For \( 2x = n\pi \), solve to find \( x = \frac{n\pi}{2} \).

Within the problem's domain \([-2\pi, 2\pi]\), \( n \) takes values \(-4, -3, -2, -1, 0, 1, 2, 3, 4\), yielding \( x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \).

Algebraically deriving these values ensures a precise solution set for intersections. It also helps verify the graphical approximation, reinforcing learning through dual perspectives – visual and algebraic.

Trigonometric Functions

Understanding the basis of trigonometric functions like sine, which repeat in cycles, can greatly illuminate the exploration of their intersections. The functions examined, \( f(x) = \sin(2x) + 1 \) and \( g(x) = 2\sin(2x) + 1 \), illustrate typical modifications:

• The argument \( 2x \) affects the function's frequency, increasing periodicity relative to \( \sin(x) \).
• The addition of constant values results in vertical shifts, where \(+1\) raises the entire sine curve.
• Vertical stretching, illustrated by \( 2\sin(2x) \), influences amplitude, enlarging peak and trough values.

Each transformation impacts the waveform behavior and, consequently, the intersection outcomes. Recognizing these characteristics provides a structured approach to predict behaviors and detect intersection points, solidifying foundational trigonometric concepts linked to frequency, amplitude, and phase shift.