Problem 61 - Calculus: Early Transcendentals

Question

Prove, without graphing, that the graph of the function has at least two \( x \)-intercepts in the specified interval. \( y = \sin x^3 \), \( (1, 2) \)

Step-by-Step Solution

Answer

The function has at least two x-intercepts in (1, 2) because \( x = (\pi)^{1/3} \) and \( x = (2\pi)^{1/3} \) are solutions.

Step 1: Understanding the problem

We need to show that the function \( y = \sin(x^3) \) crosses the x-axis at least twice in the interval \( (1, 2) \). This means we need to find at least two different \( x \) values in the interval where \( \sin(x^3) = 0 \).

Step 2: Identifying the properties of the function

The function \( y = \sin(x^3) \) is the sine of \( x\) cubed. The sine function is periodic and has zeros whenever its argument is a multiple of \( \pi \): \( \sin x = 0 \to x = n\pi \). For \( y = \sin(x^3) \), we need \( x^3 = n\pi \) for \( n \) being an integer.

Step 3: Checking the range of \( x^3 \) in \( (1, 2) \)

Let's consider the cube of the boundary values. \( 1^3 = 1 \) and \( 2^3 = 8 \). Hence, as \( x \) varies from 1 to 2, \( x^3 \) varies from 1 to 8. We need to find integers \( n \) such that \( n\pi \) falls between 1 and 8.

Step 4: Finding multiples of \( \pi \) within the interval

We need to find integer values of \( n \) for which \( 1 < n \pi < 8 \). Since \( \pi \) is approximately 3.14, we find that for \( n=1, \pi \approx 3.14 \) and for \( n=2, 2\pi \approx 6.28 \). Both values are within the range of 1 to 8.

Step 5: Concluding the proof

Thus, the function \( y = \sin(x^3) \) has zeros at \( x^3 = \pi \) and \( x^3 = 2\pi \) in the interval. Solving for \( x \) gives \( x = (\pi)^{1/3} \) and \( x = (2\pi)^{1/3} \), both of which are within \( (1, 2) \). Therefore, there are at least two \( x \)-intercepts in the interval.

Key Concepts

Sine Function

The sine function is a key trigonometric function denoted as \( \sin(x) \). It describes the ratio of the opposite side to the hypotenuse in a right triangle. Importantly, the sine function is periodic, meaning it repeats its values in a regular pattern. This periodicity can be observed with a period of \( 2\pi \).
The graph of the sine function waves above and below the x-axis. Its peaks reach a maximum value of 1 and its valleys reach a minimum value of -1.
When analyzing functions like \( y = \sin(x^3) \), it's crucial to remember that the sine function will be zero whenever its argument is a multiple of \( \pi \). This property is essential for finding x-intercepts, as it allows you to determine when the function crosses the x-axis.

Interval Analysis

Interval analysis involves examining a specific range of input values, or domain, to understand a function's behavior within that segment. This technique helps determine characteristics such as where a function is positive, negative, or crosses the x-axis.
In the problem of finding x-intercepts for \( y = \sin(x^3) \) over the interval \( (1, 2) \), interval analysis guides us in assessing how \( x^3 \) changes as \( x \) progresses from 1 to 2. Within this interval, \( x^3 \) will vary from 1 to 8.
Analyzing the interval helps identify the specific points at which \( x^3 \) equals integer multiples of \( \pi \), thereby indicating potential zeros of the sine function.

Zeros of Trigonometric Functions

Zeros of trigonometric functions correspond to values where the function evaluates to zero. For the sine function, these occur at integer multiples of \( \pi \) (i.e., \( n\pi \)). Determining zeros of a trigonometric function within a given interval gives insights into where the function intersects the x-axis.
For the function \( y = \sin(x^3) \), one must solve the equation \( x^3 = n\pi \) to find zeros. Converting these back to \( x \) by determining \( x = (n\pi)^{1/3} \), you locate the specific x-intercepts that fall within the desired interval. Each zero found signifies an x-intercept where the sine function changes sign.
In the interval \( (1, 2) \), zeros at \( x = (\pi)^{1/3} \) and \( x = (2\pi)^{1/3} \) ensure that the function crosses the x-axis at least twice.

Proof Techniques in Calculus

Proof techniques in calculus often involve verifying certain properties mathematically without graphing or using intuitive assumptions. This problem requires a non-graphical proof to show the presence of x-intercepts.
The key to proving the existence of x-intercepts in \( y = \sin(x^3) \) involves:

• Understanding the behavior of the sine function.
• Performing interval analysis to predict how \( x^3 \) changes.
• Calculating when \( x^3 = n\pi \) for integers \( n \).
• Determining values of \( x \) from \( x^3 \) that fall within the interval.

Through these logical steps, you can validate that at least two x-intercepts exist in the interval \( (1, 2) \). Using calculus principles ensures accuracy and a deeper understanding of the underlying mathematical relationships without relying on visual aids.