Problem 49 - College Algebra with Corequisite Support

Question

For the following exercises, evaluate each function. Round answers to four decimal places, if necessary. \(f(x)=1.2 e^{2 x}-0.3,\) for \(f(3)\)

Step-by-Step Solution

Answer

The value of \(f(3)\) is approximately 483.8146.

Step 1: Substitute the Value

First, substitute \(x=3\) into the function. The function given is \(f(x) = 1.2 e^{2x} - 0.3\). After substitution, it becomes \(f(3) = 1.2 e^{2 \times 3} - 0.3\).

Step 2: Simplify the Exponent

Calculate \(2 \times 3\) to simplify the exponent. This results in \(f(3) = 1.2 e^{6} - 0.3\).

Step 3: Calculate the Exponential Function

Evaluate \(e^{6}\). Using a calculator, \(e^{6} \approx 403.4288\).

Step 4: Multiply by the Coefficient

Multiply the result of the exponential function by 1.2: \(1.2 \times 403.4288 = 484.1146\).

Step 5: Subtract the Constant

Finally, subtract the constant term 0.3 from the previous result: \(484.1146 - 0.3 = 483.8146\).

Step 6: Round the Answer

Round the final result to four decimal places, giving \(f(3) = 483.8146\).

Key Concepts

Rounding Decimal Places

When working with numbers, especially in functions or calculations, you often need to round decimal numbers to a certain number of places. This makes the numbers easier to read and interpret without losing too much accuracy. Rounding to four decimal places means keeping four numbers after the decimal point. For example, 483.81462 would round to 483.8146.

• Locate the digit at the fourth decimal place.
• Check the digit immediately following it. If it is 5 or greater, increase the fourth decimal place by one. If it's less than 5, keep it the same.
• Drop all digits to the right of this position.

Rounding helps in expressing complex numbers more simply without significantly changing the value, which is crucial when precision isn't critical in problem-solving.

Function Substitution

Function substitution is the process of replacing a variable in an equation or function with a specific value or another expression. It's a fundamental technique in evaluating expressions and solving equations.
When the exercise requires evaluating the function for a specific number, such as finding \( f(3) \) for the function \( f(x) = 1.2e^{2x} - 0.3 \), you substitute \( x = 3 \) into the function.

• Replace every instance of the variable \( x \) with 3, the provided value.
• Recalculate the function with this new value to understand its effect.
• The performance of this evaluation provides the specific output for that input.

Substitution is essential for understanding how functions behave and are used in solving real-world problems and simplifying expressions.

Exponential Growth

Exponential growth occurs when the rate of increase of a quantity is proportional to its current value. It is a powerful concept often described by equations involving the mathematical constant \( e \), which is approximately 2.71828. This constant appears in many mathematical contexts, particularly in calculus and complex calculations.
In the example function \( f(x) = 1.2e^{2x} - 0.3 \), \( e^{2x} \) represents exponential growth because \( e \) is raised to the power related to \( x \). Exponential functions grow rapidly. When \( x \) increases, the exponential term forms a steep upward curve.

• This type of function is encountered in fields such as finance for compounding interest, in biology for population growth models, and in physics for radioactive decay.
• Exponential growth distinguishes itself from linear growth with its nonlinear, accelerating nature.

In real-life scenarios, understanding exponential growth is crucial for predicting future outcomes based on current inputs.

Algebraic Operations

Algebraic operations include processes such as addition, subtraction, multiplication, and division, applied to numbers and algebraic expressions. They form the basis for solving equations and simplifying expressions.
In the solution provided, various algebraic operations are applied to find \( f(3) \):

• Multiply \( e^{6} \) by 1.2, which involves scaling the exponential term by the coefficient.
• Subtract 0.3 from the result to adjust the final value by a constant amount.

These operations transform functions, allowing calculation of precise values from complex expressions.

They help break down complicated problems into manageable steps, making it easier to understand and compute expressions in algebra.