Math Problem Statement

the graph of f(x)=-tan x for x between -π/2 and π/2?

Solution

The graph of the function f(x)=tan(x)f(x) = -\tan(x) for xx in the interval (π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) has some distinctive characteristics. Here's a detailed description:

Key Features of the Graph

  1. Asymptotes:

    • The function f(x)=tan(x)f(x) = -\tan(x) has vertical asymptotes at x=π2x = -\frac{\pi}{2} and x=π2x = \frac{\pi}{2}. The graph approaches these lines but never touches them.
  2. Behavior of the Function:

    • The tangent function, tan(x)\tan(x), increases from negative infinity to positive infinity as xx moves from π2-\frac{\pi}{2} to π2\frac{\pi}{2}. However, since the function is negated, f(x)=tan(x)f(x) = -\tan(x) will decrease from positive infinity to negative infinity as xx increases from π2-\frac{\pi}{2} to π2\frac{\pi}{2}.
  3. Symmetry:

    • The graph is symmetric with respect to the origin, meaning it has odd symmetry. This means f(x)=f(x)f(-x) = -f(x).
  4. Intersections:

    • The graph intersects the origin at the point (0,0)(0, 0) because tan(0)=0\tan(0) = 0, so f(0)=0f(0) = 0.

Description of the Graph

  • Starting at x=π2x = -\frac{\pi}{2}: The graph starts near positive infinity because tan(x)\tan(x) approaches negative infinity as xx approaches π2-\frac{\pi}{2}, and the negative sign inverts this.

  • Passing through the Origin (0,0)(0, 0): The graph crosses the origin because f(0)=tan(0)=0f(0) = -\tan(0) = 0.

  • Ending at x=π2x = \frac{\pi}{2}: The graph ends near negative infinity as xx approaches π2\frac{\pi}{2} because tan(x)\tan(x) approaches positive infinity, and the negative sign inverts this.

If you want a visual graph, I can describe how to create it or provide more details. Would you like further details or a visual representation?


Here are five related questions that might interest you:

  1. How does the graph of f(x)=tan(x)f(x) = \tan(x) differ from f(x)=tan(x)f(x) = -\tan(x) in the same interval?
  2. What happens to the graph of f(x)=tan(x)f(x) = -\tan(x) if the interval is extended beyond π2-\frac{\pi}{2} and π2\frac{\pi}{2}?
  3. How would the graph change if we considered f(x)=tan(x)f(x) = \tan(x) within a different interval, such as (π2,3π2)\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)?
  4. What is the effect of adding a constant to f(x)=tan(x)f(x) = -\tan(x), such as f(x)=tan(x)+2f(x) = -\tan(x) + 2?
  5. How do the graphs of the functions f(x)=sin(x)f(x) = \sin(x), f(x)=cos(x)f(x) = \cos(x), and f(x)=tan(x)f(x) = \tan(x) compare within the interval (π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)?

Tip: When graphing trigonometric functions, paying attention to the behavior near asymptotes and zeros helps in understanding the overall shape and transformation effects on the function.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Graphing trigonometric functions

Formulas

\( f(x) = -\tan(x) \)

Theorems

Properties of trigonometric functions
Symmetry of functions

Suitable Grade Level

Advanced High School