Problem 16
Question
Use \(f(x)=\frac{10}{1+2 e^{-0.3 x}}\) Evaluate \(f(1)\).
Step-by-Step Solution
Verified Answer
Use a calculator in the last step to get the approximate value for \(f(1)\).
1Step 1: Substitution
Substitute \(x = 1\) into the given function \(f(x)=\frac{10}{1+2 e^{-0.3 x}}\). This gives us \(f(1)=\frac{10}{1+2 e^{-0.3(1)}}\).
2Step 2: Simplify the exponent
Simplify the expression in the exponent \(-0.3(1) = -0.3\). Thus, \(f(1)=\frac{10}{1+2 e^{-0.3}}\).
3Step 3: 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 \(f(1)\).
Key Concepts
Function EvaluationSubstitution MethodExponential Decay
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.
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.
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.
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.
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.
Other exercises in this chapter
Problem 15
Verify that the given functions are inverses of each other. $$f(x)=x^{3}+2 ; g(x)=\sqrt[3]{x-2}$$
View solution Problem 15
Evaluate each expression to four decimal places using a calculator. $$e^{-2.5}$$
View solution Problem 16
Solve the exponential equation. Round to three decimal places, when needed. $$6\left(0.9^{x}\right)=7$$
View solution Problem 16
In Exercises \(15-20,\) use the properties of logarithms to simplify each expression by eliminating all exponents and radicals. Assume that \(x, y > 0\). $$\log
View solution