Math Problem Statement

( C(x) = \frac{100}{1 + e^{-0.5(x - 10)}} ) between ( x = 0 ) and ( x = 20 )

Solution

The function given is:

C(x)=1001+e0.5(x10)C(x) = \frac{100}{1 + e^{-0.5(x - 10)}}

This is a sigmoid function, often used to model growth processes or probability distributions.

Analysis of the Function:

  1. Domain: The function is defined for all xx, but you're interested in the interval x=0x = 0 to x=20x = 20.

  2. Behavior:

    • As xx approaches 0, C(x)C(x) will start near 0.
    • As xx increases, the function gradually increases and approaches 100 asymptotically.
  3. Key points:

    • Midpoint: At x=10x = 10, the exponent 0.5(x10)-0.5(x - 10) becomes 0, so C(10)=1001+e0=1002=50C(10) = \frac{100}{1 + e^0} = \frac{100}{2} = 50.
    • Growth rate: The rate at which C(x)C(x) approaches 100 is controlled by the coefficient 0.5 in the exponent.
  4. Values at Specific Points:

    • C(0)=1001+e5C(0) = \frac{100}{1 + e^{5}}
    • C(20)=1001+e5C(20) = \frac{100}{1 + e^{-5}}

Would you like me to compute specific values or plot this function for better visualization?

Related Questions:

  1. What is the value of C(x)C(x) at x=0x = 0, x=5x = 5, x=10x = 10, x=15x = 15, and x=20x = 20?
  2. How does changing the coefficient in the exponent (e.g., from 0.5 to 1) affect the shape of the sigmoid curve?
  3. What happens to C(x)C(x) if we extend the interval beyond x=20x = 20?
  4. How does the midpoint x=10x = 10 relate to the function's inflection point?
  5. What are some real-world applications of this sigmoid function?

Tip: Sigmoid functions like this one are often used in logistic regression models and neural networks to map predictions to probabilities.

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Math Problem Analysis

Mathematical Concepts

Functions
Exponential Functions
Logistic Functions

Formulas

Logistic function formula C(x) = 100 / (1 + e^(-0.5(x - 10)))

Theorems

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Suitable Grade Level

Advanced High School and Beyond