Math Problem Statement

g(x)=sec x [0,2π]

Solution

To find the behavior of the function g(x)=sec(x)g(x) = \sec(x) over the interval [0,2π][0, 2\pi], let's break it down:

1. Definition of sec(x)\sec(x):

The secant function is defined as: g(x)=sec(x)=1cos(x)g(x) = \sec(x) = \frac{1}{\cos(x)} This function is undefined when cos(x)=0\cos(x) = 0, because division by zero is undefined.

2. Points where sec(x)\sec(x) is undefined:

The cosine function equals zero at the following points in the interval [0,2π][0, 2\pi]: x=π2,3π2x = \frac{\pi}{2}, \, \frac{3\pi}{2} Thus, g(x)=sec(x)g(x) = \sec(x) has vertical asymptotes at x=π2x = \frac{\pi}{2} and x=3π2x = \frac{3\pi}{2}.

3. Behavior of sec(x)\sec(x) between critical points:

  • In the interval [0,π2)[0, \frac{\pi}{2}), cos(x)>0\cos(x) > 0, so sec(x)>0\sec(x) > 0.
  • In the interval (π2,π](\frac{\pi}{2}, \pi], cos(x)<0\cos(x) < 0, so sec(x)<0\sec(x) < 0.
  • In the interval [π,3π2)[\pi, \frac{3\pi}{2}), cos(x)<0\cos(x) < 0, so sec(x)<0\sec(x) < 0.
  • In the interval (3π2,2π](\frac{3\pi}{2}, 2\pi], cos(x)>0\cos(x) > 0, so sec(x)>0\sec(x) > 0.

4. Range of g(x)g(x):

Since sec(x)\sec(x) is the reciprocal of cos(x)\cos(x), and 1cos(x)1-1 \leq \cos(x) \leq 1, the range of sec(x)\sec(x) is: (,1][1,)(-\infty, -1] \cup [1, \infty) This means that sec(x)\sec(x) never takes values between -1 and 1.

5. Graphical behavior:

  • There are vertical asymptotes at x=π2x = \frac{\pi}{2} and x=3π2x = \frac{3\pi}{2}.
  • The graph alternates between positive and negative values, going to ++\infty near x=0x = 0, π\pi, and 2π2\pi, and -\infty near π2\frac{\pi}{2} and 3π2\frac{3\pi}{2}.

Would you like further details on any of these points or a specific graph of the function?


Here are 5 related questions to expand on this:

  1. What is the domain of the secant function?
  2. How do you find vertical asymptotes for trigonometric functions?
  3. What are the key properties of sec(x)\sec(x) and how do they compare with cos(x)\cos(x)?
  4. How can you determine intervals of increase and decrease for sec(x)\sec(x)?
  5. How do other reciprocal trigonometric functions behave over [0,2π][0, 2\pi]?

Tip: The secant function tends to infinity near the points where cos(x)\cos(x) approaches zero. Understanding this behavior helps in identifying its vertical asymptotes.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Secant Function
Behavior of Trigonometric Functions

Formulas

g(x) = sec(x) = 1 / cos(x)

Theorems

Reciprocal Trigonometric Function
Vertical Asymptotes in Trigonometric Functions

Suitable Grade Level

Grades 10-12