Question

Horizontal Tangent Line Determine the point(s) in the interval \((0,2 \pi)\) at which the graph of 4$f(x)=2 \cos x+\sin 2 x$$ has a horizontal tangent.

Short answer

The points where the graph of \(f(x) = 2 \cos x + \sin 2x\) has a horizontal tangent within the interval \((0, 2\pi)\) are approximately \(x \approx 0.900\) and \(x \approx 4.068\).

Step-by-step solution

Compute derivative of the function

The first derivative of the function \(f(x) = 2 \cos x + \sin 2x\) using the chain rule is: \(f'(x)=-2\sin x+ 2\cos2x\).

Set the derivative to zero

To find the x-coordinates where the function has a horizontal tangent, solve the equation \(f'(x) = 0\) thus: \(-2\sin x+ 2\cos2x = 0\).

Solve the equation for \(x\)

Rearranging the terms, we get \(tanx = cos2x\). Solving this equation within the interval \((0, 2\pi)\) will require graphing or a numerical method like the Newton-Raphson method. Here, approximation by a solver may be used to find that \(x \approx 0.900\) and \(x \approx 4.068\) are the solutions.

Check that the solutions are within the interval

The solutions \(x \approx 0.900\) and \(x \approx 4.068\) both lie within the interval \((0, 2\pi)\), thus they are valid solutions.

Key concepts

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent play a crucial role in calculus, especially when dealing with periodic functions. In this exercise, we encounter two primary trigonometric functions:

• The cosine function, denoted as \( \cos x \), which describes the x-coordinate of a point on the unit circle as it rotates through the angle \( x \).
• The sine function, represented as \( \sin 2x \), highlighting a periodic function with twice the frequency of \( \sin x \). This modifies the standard sine wave to complete its cycle twice as fast.

These functions are periodic with a period of \( 2\pi \), meaning their values repeat every \( 2\pi \) radians. Understanding the behavior of these functions is vital for solving problems involving wave-like patterns or rotational motion. Recognizing their graphical representation can assist in visualizing the problem areas and potential points of interest, such as where a function might attain a maximum or cross a zero line.

Derivative Calculation

Derivatives indicate how a function changes or how it slants at any given point. For our function \( f(x) = 2 \cos x + \sin 2x \), finding the derivative helps identify where the slope of the tangent to the curve is horizontal. This implies zero slope, and hence, a horizontal tangent line. The derivative calculation steps include:

• The derivative of \( \cos x \) is \( -\sin x \).
• By using the chain rule, the derivative of \( \sin 2x \) becomes \( 2\cos 2x \), because of the composition with the linear function inside the sine function.

Thus, we can derive \( f'(x) = -2 \sin x + 2 \cos 2x \). This expression represents the rate of change of the original function and is critical for finding points of horizontal tangency.

Chain Rule

The chain rule is essential when dealing with composite functions, which involve functions within other functions. In our case, \( \sin 2x \) is a composite function, comprised of the outer sine function applied to the inner expression \( 2x \). The chain rule states that to differentiate a composition of functions, you multiply the derivative of the outer function by the derivative of the inner function.

For the derivative of \( \sin 2x \):

• First, think of \( \sin u \) where \( u = 2x \). The derivative of \( \sin u \) is \( \cos u \).
• Next, differentiate \( u = 2x \) to get \( \frac{du}{dx} = 2 \).
• By combining these, the result is \( 2 \cos 2x \).

This usage of the chain rule simplifies differentiating more complex, layered functions without having to manipulate them into simpler parts.

Solving Equations Numerically

When finding where a derivative equals zero for non-trivial equations, numerical methods become crucial. Our task was to solve \( -2 \sin x + 2 \cos 2x = 0 \). Many equations like this one do not have algebraic solutions that can be neatly solved by hand, so numerical methods are often employed. Two common methods include:

• Using graphs to get a visual approximation of where functions intersect or cross the horizontal axis. This is useful for seeing rough estimates that guide further analysis.
• Numerical methods like the Newton-Raphson involve iterative processes that converge towards the solution, offering precise values for where a function equals zero. This method requires an initial guess and refines it using calculus-based iterations.

Applying these methods, we determine the approximate x-values that result in the function having a horizontal tangent: \( x \approx 0.900 \) and \( x \approx 4.068 \). These solutions are validated by their presence within our specified interval of \( (0, 2\pi) \).