/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Use \(f(x)=\frac{10}{1+2 e^{-0.3... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use \(f(x)=\frac{10}{1+2 e^{-0.3 x}}\) Evaluate \(f(1)\).

Short Answer

Expert verified
Use a calculator in the last step to get the approximate value for \(f(1)\).

Step by step solution

01

Substitution

Substitute \(x = 1\) into the given function \(f(x)=\frac{10}{1+2 e^{-0.3 x}}\). This gives us \(f(1)=\frac{10}{1+2 e^{-0.3(1)}}\).
02

Simplify the exponent

Simplify the expression in the exponent \(-0.3(1) = -0.3\). Thus, \(f(1)=\frac{10}{1+2 e^{-0.3}}\).
03

Calculate the value

Now calculate the value of \(e^{-0.3}\) using a calculator to find an approximate value. Then, plug this value into the previous equation to find the value of \(f(1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is a crucial process in mathematics, allowing us to determine the output of a function for a specific input. This is akin to plugging a number into a function's "recipe" to see what result it produces.
When we evaluate a function, we replace the variable with the given number. For the function provided, \( f(x) = \frac{10}{1 + 2e^{-0.3x}} \), the task was to evaluate it at \( x = 1 \).
This means replacing \( x \) with 1 in every instance. We move from a general expression to a specific value, simplifying the understanding of how the function behaves for any particular input. Doing this repeatedly with different inputs can help us grasp the overall behavior of the function more completely.
Substitution Method
The substitution method is a straightforward technique to tackle the evaluation of functions. This involves replacing a variable with a specific value, transforming a general formula into a more manageable computation.
In our example, we began by substituting \( x = 1 \) into \( f(x) = \frac{10}{1 + 2e^{-0.3x}} \). This resulted in \( f(1) = \frac{10}{1 + 2e^{-0.3(1)}} \).
This method simplifies the function for a specific situation, requiring straightforward arithmetic to reach a conclusion.
  • It’s a crucial skill for solving many types of math problems.
  • It helps in evaluating functions both in exercises and real-life applications.
The substitution method ensures that we proceed step-by-step, keeping our calculations accurate.
Exponential Decay
Exponential functions describe processes that increase or decrease rapidly. An exponential decay denotes a process that reduces over time. In our function \( f(x) = \frac{10}{1 + 2e^{-0.3x}} \), the term \( e^{-0.3x} \) represents exponential decay.
Here, the base \( e \) is an irrational number approximately equal to 2.71828, and the negative exponent indicates a reduction.
  • As \( x \) increases, \( e^{-0.3x} \) becomes smaller, implying that the influence of the exponential term declines.
  • The coefficient in front of the exponential term adjusts the rate of this decay.
Understanding exponential decay involves recognizing that while values decrease, they never reach zero, modeling processes like cooling temperature or radioactive decay effectively. Hence, these concepts are applied in functions to depict real-world diminishing trends. Such insight can help predict outcomes in various fields, including finance, physics, and environmental science.

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Most popular questions from this chapter

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