Problem 13 - Algebra and Trigonometry

Question

Evaluate the function at the indicated values. $$ \begin{array}{l}{f(x)=2 x+1} \\ {f(1), f(-2), f\left(\frac{1}{2}\right), f(a), f(-a), f(a+b)}\end{array} $$

Step-by-Step Solution

Answer

f(1)=3, f(-2)=-3, f(\frac{1}{2})=2, f(a)=2a+1, f(-a)=-2a+1, f(a+b)=2a+2b+1.

Step 1: Evaluate f(1)

For \( f(1) \), substitute \( x = 1 \) into the function. \[f(x) = 2(1) + 1 = 2 + 1 = 3\]Thus, \( f(1) = 3 \).

Step 2: Evaluate f(-2)

For \( f(-2) \), substitute \( x = -2 \) into the function. \[f(x) = 2(-2) + 1 = -4 + 1 = -3\]Thus, \( f(-2) = -3 \).

Step 3: Evaluate f(\(\frac{1}{2}\))

For \( f\left(\frac{1}{2}\right) \), substitute \( x = \frac{1}{2} \) into the function. \[f(x) = 2\left(\frac{1}{2}\right) + 1 = 1 + 1 = 2\]Thus, \( f\left(\frac{1}{2}\right) = 2 \).

Step 4: Evaluate f(a)

For \( f(a) \), substitute \( x = a \) into the function. \[f(x) = 2a + 1\]Thus, \( f(a) = 2a + 1 \).

Step 5: Evaluate f(-a)

For \( f(-a) \), substitute \( x = -a \) into the function. \[f(x) = 2(-a) + 1 = -2a + 1\]Thus, \( f(-a) = -2a + 1 \).

Step 6: Evaluate f(a+b)

For \( f(a+b) \), substitute \( x = a+b \) into the function. \[f(x) = 2(a+b) + 1 = 2a + 2b + 1\]Thus, \( f(a+b) = 2a + 2b + 1 \).

Key Concepts

Substitution in Functions

Substitution in functions is a fundamental concept in mathematics. It allows us to find the output of a function when given specific inputs. Imagine functions as machines where you feed in a number, and it gives you back another number based on its rule. The act of 'feeding' the number into the function involves substituting the given integer, fraction, or variable value for the 'x' in the function expression.

For example, when we evaluate the function given as \( f(x) = 2x + 1 \), we substitute the specified values for \( x \). If we want to find \( f(1) \), we replace \( x \) with 1. The function becomes \( 2(1) + 1 \), which equals 3.

Key Points:

• Identify the value to substitute for \( x \).
• Replace every occurrence of \( x \) in the function with the identified value.
• Perform the arithmetic operations to find the result.

Understanding how to substitute values into functions will significantly help in evaluating and analyzing any type of function.

Linear Functions

Linear functions are among the simplest types of functions. They are called "linear" because their graphs form straight lines. In algebra, a linear function can generally be expressed as \( f(x) = mx + b \), where \( m \) and \( b \) are constants.

In the exercise we have \( f(x) = 2x + 1 \), which is a linear function where the slope \( m = 2 \), and the y-intercept \( b = 1 \). This means the line crosses the y-axis at 1 and rises two units vertically for every unit it moves horizontally to the right.

Characteristics of Linear Functions:

• Graphically represented by a straight line.
• Has a constant rate of change (slope).
• No variables are multiplied together or taken to a power other than one.

Linear functions form the foundation for understanding more complex mathematical functions and are frequently used to model real-world situations.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and at least one arithmetic operation (like addition or multiplication). In our function \( f(x) = 2x + 1 \), \( 2x + 1 \) is an algebraic expression.

Breaking it down:

• \( 2x \) consists of a coefficient 2 and a variable \( x \).
• The term 1 is a constant.
• The expression uses addition to combine terms.

Understanding algebraic expressions is crucial because they form the building blocks of functions. When substituting in different values, the form of algebraic expressions helps determine the output by simplifying to a single value or expression.