/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Do the graphs of the functions i... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 40

Do the graphs of the functions in Exercises \(39-42\) have any horizontal tangents in the interval \(0 \leq x \leq 2 \pi ?\) If so, where? If not, why not? Visualize your findings by graphing the functions with a grapher. $$ y=2 x+\sin x $$

Short Answer

Expert verified
No, the function has no horizontal tangents in the interval \(0 \leq x \leq 2\pi\).

Step by step solution

01

Find the derivative of the function

To find the horizontal tangents, we first need to find the derivative of the given function. The function given is \( y = 2x + \sin x \). The derivative, \( y' \), is calculated as follows: \[ y' = \frac{d}{dx}(2x + \sin x) = 2 + \cos x \]
02

Set the derivative equal to zero

Horizontal tangents occur where the derivative equals zero, as this indicates a slope of zero. Thus, we set the derivative equal to zero: \[ 2 + \cos x = 0 \] This implies that \( \cos x = -2 \).
03

Analyze feasibility in the interval

Check whether \( \cos x = -2 \) is possible. The range of \( \cos x \) is \([-1, 1]\). Since \( -2 \) is outside this range, the equation \( \cos x = -2 \) has no solutions in real numbers.
04

Conclusion on horizontal tangents

Since \( \cos x = -2 \) has no solution within the range of cosine in real numbers and within the given interval \(0 \leq x \leq 2\pi\), there are no points where the derivative is zero. Hence, the function has no horizontal tangents in the specified interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative Calculation
Understanding the derivative is crucial for identifying horizontal tangents. A derivative represents the slope of a function at any given point. To find horizontal tangents, we look for places where this slope is zero, meaning the function flattens out. For the function \( y = 2x + \sin x \), the derivative is calculated by differentiating each term separately:
  • The derivative of \( 2x \) is 2, since the derivative of a linear function \( ax \) is simply the constant \( a \).
  • For the \( \sin x \) term, the derivative is \( \cos x \), as the derivative of \( \sin x \) with respect to \( x \) is \( \cos x \).
Putting it together, the derivative \( y' \) is \( 2 + \cos x \). This equation represents how the slope of the function changes across the input values.
Interval Analysis
The interval \(0 \leq x \leq 2\pi\) is crucial because it outlines the range in which we need to check for horizontal tangents. With the interval defined, our task is to see if there exists any \( x \) within this range where the derivative equals zero鈥攖his is what signifies a horizontal tangent.When we set the derivative \( 2 + \cos x = 0 \) to find such points and solve for \( \cos x \), we get \( \cos x = -2 \). However, a closer look at trigonometric functions reveals that the cosine function has a range of \([-1, 1]\). Therefore, it's impossible for \( \cos x = -2 \) to hold true, meaning there's no point within the interval where the function has a horizontal slope.
Trigonometric Functions
Trigonometric functions like \( \sin x \) and \( \cos x \) are periodic, with known ranges and behaviors. The range of \( \cos x \) is \([-1, 1]\), a fundamental property that plays a pivotal role in our analysis here. When solving \( 2 + \cos x = 0 \), the solution implies \( \cos x = -2 \), but \( \cos x \) never actually reaches -2 with any real \( x \).
  • This discrepancy arises from trying to equate \( \cos x \), which oscillates between -1 and 1, to a value outside its maximum and minimum.
  • Recognizing the limitations of trigonometric functions is essential for accurately identifying potential solutions or confirming their impossibility.
Knowing these characteristics helps us understand why no real \( x \) can satisfy our derivative equation in the given interval.
Graphical Visualization
Visualizing the function \( y = 2x + \sin x \) can be a powerful tool for understanding the behavior of the function and confirming theoretical findings. Graphs provide a visual representation of mathematical concepts, such as slopes and tangents.
  • By graphing \( y = 2x + \sin x \), we observe how the function behaves over the interval \( 0 \leq x \leq 2\pi \).
  • We can see the continuous rise due to the linear term \( 2x \), alongside the undulating effect added by \( \sin x \).
  • No flat spots or horizontal tangents appear in the graph, visually supporting the analytic conclusion derived from the derivative.
Thus, graphical representation not only confirms our mathematical analysis but also enhances our intuitive understanding of the function's behavior over the given range.

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