Problem 44 - College Algebra with Modeling and Visualization

Question

Approximate \(f(x)\) to four decimal places. $$ f(x)=4\left(e^{-0.3 x}-e^{-0.6 x}\right), \quad x=1.6 $$

Step-by-Step Solution

Answer

The approximation of \( f(1.6) \) is 0.9412.

Step 1: Identify the function and given value of x

The function to approximate is \( f(x) = 4(e^{-0.3x} - e^{-0.6x}) \). The value given for \( x \) is 1.6.

Step 2: Substitute the value of x into the function

Replace \( x \) in the function with 1.6. This gives us:\[ f(1.6) = 4(e^{-0.3(1.6)} - e^{-0.6(1.6)}) \]

Step 3: Calculate each exponential component

First, compute \( e^{-0.3(1.6)} \):\[ e^{-0.3 imes 1.6} = e^{-0.48} \approx 0.6188 \] Next, compute \( e^{-0.6(1.6)} \):\[ e^{-0.6 imes 1.6} = e^{-0.96} \approx 0.3835 \]

Step 4: Determine the difference of exponentials

Subtract the two results calculated in Step 3:\[ e^{-0.48} - e^{-0.96} = 0.6188 - 0.3835 = 0.2353 \]

Step 5: Multiply by 4 to find the approximation

Multiply the result from Step 4 by 4 to get:\[ f(1.6) = 4 imes 0.2353 = 0.9412 \]

Step 6: Approximate to four decimal places

Since the previous step already provides the result to four decimal places, the approximation is complete: \( f(1.6) \approx 0.9412 \).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable power. In the expression \( f(x) = 4(e^{-0.3x} - e^{-0.6x}) \), the base \( e \) is a constant approximately equal to 2.71828. Exponential functions are widely used in various fields including biology, finance, and physics, mostly due to their ability to model growth and decay processes. Here, we have an expression with two exponential components, \( e^{-0.3x} \) and \( e^{-0.6x} \), representing decay.
Since \( e \) is the natural logarithm base, when raised to a negative power, it represents a decrease over time or space. This is why values of \( e^{-0.3x} \) and \( e^{-0.6x} \) are smaller than 1, illustrating decay. In essence:

• As \( x \) increases, \( e^{-0.3x} \) and \( e^{-0.6x} \) get smaller.
• This makes the difference \( e^{-0.3x} - e^{-0.6x} \) indicative of how one component decays faster than the other.

The coefficient of 4 scales the resulting value, emphasizing the magnitude of decay difference derived from the function.

Numerical Methods

Numerical methods are mathematical tools used to find approximate solutions to complex mathematical problems. In scenarios involving transcendental functions like exponentials, computing exact values is often impractical. Hence, we employ numerical techniques.
Here, the numerical approximation involves substituting a specific value of \( x \) to evaluate the function \( f(x) = 4(e^{-0.3x} - e^{-0.6x}) \) for \( x = 1.6 \). Key steps include:

• Substituting \( x \) into each exponential component, calculating their values using approximation methods for exponentials, such as series or calculators.
• Computing the difference between the two radically changed values of exponentials.
• Multiplying the difference by 4 to finalize the approximation.

Numerical methods help achieve a four-decimal precision in the solution \( f(1.6) \approx 0.9412 \), showcasing its practicality for solving real-world problems efficiently without algebraic complexity.

Algebraic Calculation

Algebraic calculation involves performing operations using symbols and formulas to derive a numerical result. Given the exponential function, we proceed by employing algebraic techniques to approximate \( f(x) \).
To do so, substitute a particular \( x \) to evaluate the expression. With \( x = 1.6 \), the calculation simplifies the complex components. Each exponential term \( e^{-0.3 \times 1.6} \) and \( e^{-0.6 \times 1.6} \) is solved individually before being subtracted, leveraging basic arithmetic operations.
The result illustrates:

• The calculation sequence of handling exponential terms separately.
• Once computed, finding the difference accentuates how arithmetic operations streamline solving equations.
• The end step multiplies by 4, enhancing magnitude with a straightforward calculation task.

Through successive algebraic simplifications, complexities reduce, converting expressions into approximate decimal values. Instructive problem-solving aligns with distinguishing components, easing understanding of exponential approximations.