Problem 20 - Precalculus with Limits

Question

In Exercises 17-28, evaluate the indicated function for \(f(x) = x^2 + 1\) and \(g(x) = x - 4\). \((f + g)(1)\)

Step-by-Step Solution

Answer

-1

Step 1: Evaluate f(1)

To evaluate this, you need to substitute \(x = 1\) in the function \(f(x)\). So, \(f(1) = (1)^2 + 1 = 1 + 1 = 2.

Step 2: Evaluate g(1)

Now, substitute \(x = 1\) into the function \(g(x)\). This gives you \(g(1) = 1 - 4 = -3.

Step 3: Compute (f + g)(1)

\((f + g)(1)\) implies calculating the sum of f(1) and g(1), which is \(2 + -3 = -1.

Key Concepts

Polynomial Functions

Polynomial functions are expressions that involve variables raised to whole number powers. They can include constants, variables, and the sum of terms such as squares and higher-degree terms. The general form of a polynomial function in one variable is:

• On the form: \( a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \)
• Where \( a_n, a_{n-1}, \ldots, a_0 \) are constants called coefficients, and \( n \) represents the degree of the polynomial.

In the given exercise, the function \( f(x) = x^2 + 1 \) is a polynomial of degree 2. This means that the highest power of the variable \( x \) is 2. Polynomial functions like this are smooth and continuous, making them easy to work with in calculus and algebra.

Function Operations

Function operations allow us to perform arithmetic operations on functions, just like we do with numbers. These operations include addition, subtraction, multiplication, and division of functions. In the given problem, we focused on the addition of two functions \( f(x) \) and \( g(x) \).

• Addition: When adding two functions, \( (f + g)(x) \), you simply add the corresponding outputs of \( f(x) \) and \( g(x) \). For example: \( (f + g)(1) = f(1) + g(1) \).
• Subtraction: Subtracting one function from another works similarly, subtracting their outputs: \( (f - g)(x) = f(x) - g(x) \).
• Multiplication: Multiply the outputs: \( (f \cdot g)(x) = f(x) \cdot g(x) \).
• Division: Divides the outputs: \( \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} \), provided \( g(x) eq 0 \).

Function operations are crucial for function evaluation, as they help to manipulate and combine functions according to problem requirements.

Algebraic Substitution

Algebraic substitution refers to replacing a variable with a specific value or another expression. This method is often used to evaluate functions at given points or simplify expressions.

• Substitution in Evaluation: To evaluate a function, substitute the given value into the function. For example, to find \( f(1) \), replace \( x \) with \( 1 \) in the function \( f(x) = x^2 + 1 \), resulting in \( f(1) = 1^2 + 1 = 2 \).
• Use in Solving Equations: Substitution helps to solve equations by replacing variables with known values or other expressions, simplifying the problem significantly.
• Simplifying Expressions: Substitution can also simplify algebraic expressions by making complex problems more manageable using known values or simpler terms.

By mastering substitution, understanding and solving algebra problems becomes easier and more intuitive. It is a valuable tool for any math student as it simplifies both evaluation and problem-solving processes.