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Chapter 3: Problem 52

If \(f(x)=\sin 2 x-\cos 2 x\) for \(0 \leq x \leq 2 \pi,\) find the \(x\) -coordinates of all points on the graph of \(f\) at which the tangent line is horizontal.

Short Answer

Expert verified
The horizontal tangent points are at \( x = \frac{3\pi}{8}, \frac{7\pi}{8}, \frac{11\pi}{8}, \text{and } \frac{15\pi}{8} \).

Step by step solution

01

Understand the Problem

We need to find the points on the graph of the function where the tangent line is horizontal. A horizontal tangent line means the slope is zero, so we need to find where the derivative of the function equals zero.
02

Differentiate the Function

First, find the derivative of the function \f(x) = \sin(2x) - \cos(2x)\Using standard differentiation rules:\[ f'(x) = \frac{d}{dx}[\sin(2x)] - \frac{d}{dx}[\cos(2x)] \]\[ f'(x) = 2\cos(2x) + 2\sin(2x) \]
03

Set the Derivative Equal to Zero

Next, set the derivative \( f'(x) \) equal to zero to find points where the slope is zero:\[ 2\cos(2x) + 2\sin(2x) = 0 \]Simplify:\[ \cos(2x) + \sin(2x) = 0 \]
04

Solve the Trigonometric Equation

Solve the equation \( \cos(2x) + \sin(2x) = 0 \) by isolating one function:\[ \cos(2x) = - \sin(2x) \]This implies:\[ \tan(2x) = -1 \]
05

Find Solutions for \( \tan(2x) = -1 \)

Solutions to \( \tan(\theta) = -1 \) occur at \( \theta = \frac{3\pi}{4} + n\pi \), where \( n \) is an integer.Set \( 2x = \frac{3\pi}{4} + n\pi \) and solve for \( x \):\[ x = \frac{3\pi}{8} + \frac{n\pi}{2} \]
06

Determine Valid \(x\) Values

Given the range \( 0 \leq x \leq 2\pi \), calculate values for different integers \( n \):- For \( n = 0 \), \( x = \frac{3\pi}{8} \)- For \( n = 1 \), \( x = \frac{3\pi}{8} + \frac{\pi}{2} = \frac{7\pi}{8} \)- For \( n = 2 \), \( x = \frac{3\pi}{8} + \pi = \frac{11\pi}{8} \)- For \( n = 3 \), \( x = \frac{3\pi}{8} + \frac{3\pi}{2} = \frac{15\pi}{8} \)All values are within the given range.

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Key Concepts

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

Horizontal Tangent
In calculus, when we talk about horizontal tangents, we're looking for points on a curve where the slope is zero. This usually means the derivative of the function equals zero at these points.
Finding horizontal tangents is essential in calculus because it often helps in understanding the behavior of a function, indicating maximums, minimums, or flat spots on the graph.
  • A horizontal tangent implies the tangent line is perfectly flat at that point.
  • The slope, or rate of change, of the curve at that point is zero.
In the given problem, our task was to determine the x-values where the function has a horizontal tangent. We did this by finding where the derivative of the function equaled zero. Thus, calculating the derivative and solving the resulting equation was crucial in locating these points.
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent play a significant role in calculus. They help describe oscillating or wave-like behaviors, which are common in many natural and engineered systems.
When dealing with trigonometric functions, it's important to remember their periodic nature. This means they repeat their values over specific intervals. For example, sine and cosine have cycles of \(2\pi\), while tangent has a cycle of \(\pi\).
  • Each trig function has specific properties and graphs that make their behavior predictable.
  • Understanding trigonometric identities can simplify solving equations involving these functions.
In this exercise, we worked with the sine and cosine functions. We differentiated these functions using the chain rule, leading to the equation \(2\cos(2x) + 2\sin(2x) = 0\). Solving trigonometric equations often requires manipulating these identities to simplify and find solutions.
Calculus Problem-Solving
Problem-solving in calculus often involves breaking down the problem into manageable steps. This is especially true for complex functions involving derivatives. Knowing the right techniques can simplify the process.
Here's a general approach:
  • Understand the Problem: Clearly define what is being asked. Is it seeking a maximum, minimum, a point of inflection, etc.?
  • Differentiate: Find the derivative, as it is vital for understanding rates of change and slopes.
  • Solve the Derivative Equation: Set the derivative equal to zero if looking for horizontal tangents or critical points.
  • Check Solutions: Verify which solutions lie within the given domain. This step is crucial to ensure the solutions make sense in the context of the problem.
In our case, we processed the function through these steps to find where the tangent was horizontal. Each step requires careful consideration and analysis for successful calculus problem-solving.

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