Math Problem Statement

What is the period of the trigonometric function shown in the graph with a minimum point at (π, -11) and a maximum point at (6/5 π, -3.7)?

Solution

To find the period of this trigonometric function, we can examine the x-coordinates of the given minimum and maximum points.

The key points provided are:

  • Minimum at (π,11)(\pi, -11)
  • Maximum at (65π,3.7)\left(\frac{6}{5} \pi, -3.7\right)

Since the function oscillates between minimum and maximum values regularly, we can infer that half of the period spans from a minimum point to the next maximum point. So, the difference in x-coordinates between these points represents half of the period of the function.

Step-by-Step Calculation

  1. Calculate Half of the Period: Half period=65ππ\text{Half period} = \frac{6}{5} \pi - \pi Simplifying this: =6π5π5=π5= \frac{6 \pi - 5 \pi}{5} = \frac{\pi}{5}

  2. Determine the Full Period: Since π5\frac{\pi}{5} is half of the period, the full period TT is: T=2×π5=2π5T = 2 \times \frac{\pi}{5} = \frac{2\pi}{5}

Final Answer

The exact period of the function is: 2π5\frac{2\pi}{5}

Would you like more details or have any questions?


Related Questions

  1. How can we determine the amplitude of the function from the graph?
  2. What type of trigonometric function might have this period?
  3. How would the equation change if the period was doubled?
  4. Could we find the phase shift of this function with the given points?
  5. What would the next minimum and maximum points be on the graph?

Tip

In periodic functions, identifying the interval between successive peaks or troughs is a reliable method to determine the period accurately.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometric Functions
Periodicity
Wave Analysis

Formulas

Period T = 2 * (difference between maximum and minimum x-coordinates)

Theorems

Periodicity of Trigonometric Functions

Suitable Grade Level

Grades 10-12