Math Problem Statement

What can you say about the properties of the function $y=\tan{x}$ when $0\leq x < \frac{\pi}{2}$. (This is purposely open ended.)

Solution

The function y=tanxy = \tan{x} exhibits a variety of interesting properties when considering the domain 0x<π20 \leq x < \frac{\pi}{2}. Let’s go over several key aspects:

1. Domain and Range:

  • Domain: The function y=tanxy = \tan{x} is defined for all real values of xx except where x=π2+nπx = \frac{\pi}{2} + n\pi for any integer nn, because the tangent function has vertical asymptotes at these points due to division by zero in the expression tanx=sinxcosx\tan{x} = \frac{\sin{x}}{\cos{x}}.

    • In the interval 0x<π20 \leq x < \frac{\pi}{2}, tanx\tan{x} is defined for all xx, with no points of discontinuity.
  • Range: As xx increases from 0 to π2\frac{\pi}{2}, tanx\tan{x} takes all values from 0 to ++\infty.

    • Specifically, limx0+tanx=0\lim_{x \to 0^+} \tan{x} = 0 and limxπ2tanx=+\lim_{x \to \frac{\pi}{2}^-} \tan{x} = +\infty.

2. Increasing Function:

  • The tangent function is increasing in the interval 0x<π20 \leq x < \frac{\pi}{2}. This can be confirmed by noting that the derivative of tanx\tan{x}, which is sec2x\sec^2{x}, is positive for all xx in this interval. Thus, as xx increases from 0 to π2\frac{\pi}{2}, the value of tanx\tan{x} also increases.

3. Asymptotic Behavior:

  • As xx approaches π2\frac{\pi}{2}, tanx\tan{x} approaches infinity. This behavior is a result of the cosine function approaching zero as xx nears π2\frac{\pi}{2}, causing the ratio sinxcosx\frac{\sin{x}}{\cos{x}} to grow without bound.

4. Symmetry:

  • The tangent function is odd, meaning that tan(x)=tanx\tan(-x) = -\tan{x}. This symmetry is important for understanding the behavior of the tangent function across different quadrants of the unit circle, but it doesn't directly impact the interval 0x<π20 \leq x < \frac{\pi}{2}.

5. Continuity:

  • y=tanxy = \tan{x} is continuous on the interval 0x<π20 \leq x < \frac{\pi}{2}, except at the endpoint x=π2x = \frac{\pi}{2}, where it has an infinite discontinuity (vertical asymptote).

6. Graph Behavior:

  • The graph of y=tanxy = \tan{x} starts at the origin (0,0) and increases without bound as xx approaches π2\frac{\pi}{2}. The graph has no maximum or minimum within this domain, only a vertical asymptote at x=π2x = \frac{\pi}{2}.
  • The graph is smooth and monotonic, meaning it has no sudden jumps or oscillations.

7. Periodicity:

  • While tanx\tan{x} is periodic with period π\pi, within the interval 0x<π20 \leq x < \frac{\pi}{2}, the function behaves as if it has no period, increasing without repeating.

Would you like more details on any of these properties? Or do you have further questions about the behavior of the function?

Related Questions:

  1. How do the derivative and second derivative of y=tanxy = \tan{x} behave for 0x<π20 \leq x < \frac{\pi}{2}?
  2. What is the relationship between y=tanxy = \tan{x} and its inverse function y=tan1xy = \tan^{-1}{x} in the given interval?
  3. Can you graph the function y=tanxy = \tan{x} over 0x<π20 \leq x < \frac{\pi}{2} and identify key features like intercepts and asymptotes?
  4. What would the limit of tanx\tan{x} be if xπ2x \to \frac{\pi}{2} from the left side?
  5. How does y=tanxy = \tan{x} compare to other trigonometric functions like y=sinxy = \sin{x} or y=cosxy = \cos{x} in terms of growth within this interval?

Tip:

The behavior of tangent near its asymptotes can be better understood by observing the function’s graph or analyzing the corresponding values of sine and cosine as xx nears π2\frac{\pi}{2}.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Tangent Function
Limits
Continuity
Asymptotes

Formulas

y = tan{x} = sin{x}/cos{x}
lim(x → 0⁺) tan{x} = 0
lim(x → π/2⁻) tan{x} = +∞
y' = sec²{x}

Theorems

Limits of trigonometric functions
Continuity of functions
Behavior near asymptotes

Suitable Grade Level

Grades 11-12