Problem 7 - Calculus Volume 1

Question

For the following exercises, find the values for each function, if they exist, then simplify. a. \(f(0)\) b. \(f(1)\) c. \(f(3)\) d. \(f(-x)\) e. \(f(a)\) f. \(f(a+h)\) $$f(x)=5 x-2$$

Step-by-Step Solution

Answer

a. -2, b. 3, c. 13, d. \(-5x - 2\), e. \(5a - 2\), f. \(5a + 5h - 2\)

Step 1: Evaluate f(0)

To find the value of the function at zero, plug in 0 for the variable x in the function \(f(x) = 5x - 2\). Thus, \(f(0) = 5(0) - 2 = -2\).

Step 2: Evaluate f(1)

Substitute 1 for x in the function \(f(x) = 5x - 2\). Therefore, \(f(1) = 5(1) - 2 = 3\).

Step 3: Evaluate f(3)

Replace x with 3 in the equation \(f(x) = 5x - 2\). Hence, \(f(3) = 5(3) - 2 = 13\).

Step 4: Evaluate f(-x)

Substitute \(-x\) for x in the function \(f(x) = 5x - 2\). This results in \(f(-x) = 5(-x) - 2 = -5x - 2\).

Step 5: Evaluate f(a)

Plug in \(a\) instead of x in the equation \(f(x) = 5x - 2\). Thus, \(f(a) = 5a - 2\).

Step 6: Evaluate f(a+h)

Replace x with \(a + h\) in the function \(f(x) = 5x - 2\). Therefore, \(f(a+h) = 5(a+h) - 2 = 5a + 5h - 2\).

Key Concepts

Linear Functions

Linear functions are foundational concepts in mathematics, particularly useful in describing relationships that change at a constant rate. The key aspect of linear functions is their simplicity. They follow the form:

• \( f(x) = mx + b \)

where \(m\) represents the slope, depicting the rate of change, and \(b\) represents the y-intercept, indicating where the line crosses the y-axis.
In the exercise, the function defined is \( f(x) = 5x - 2 \), which is linear. Here, the slope \(m\) is 5, signifying for every one-unit increase in \(x\), \(f(x)\) increases by 5 units, and the y-intercept \(b\) is -2, showing the function intersects the y-axis at \( -2 \).
This constant rate of change makes linear functions straightforward to understand, predict, and evaluate, making them significant in many real-world applications.

Substitution Method

Substitution is an effective method used to find the value of a function for given inputs. It involves replacing the variable in the function with a specific value. Let's walk through how substitution works using our example.
Suppose you need to find \(f(0)\). You begin by substituting 0 wherever you see \(x\) in the function, so the evaluation becomes:

• \( f(0) = 5(0) - 2 \)
• Simplify to get: \(-2\)

This process can be repeated for any value you want to substitute for \(x\).
Another example is \(f(1)\):

• \( f(1) = 5(1) - 2 = 3 \)

Substitution is a straightforward method, facilitating the determination of function outputs for specific inputs, whether it’s a number, a variable, or an expression.

Simplifying Expressions

Simplifying expressions is an integral step in mathematics that makes equations easier to work with by rewriting them in their simplest form. In the context of the given exercise, after substituting values, simplifying is necessary to find the clean, uncomplicated form of the function's output.
For example, when evaluating \( f(a+h) \) from the exercise:

• Substitute to get: \( f(a + h) = 5(a + h) - 2 \)
• Distribute the 5: \( 5a + 5h - 2 \)

This simplifies the expression by ensuring all like terms have been combined and any arithmetic (like distribution of numbers) has been completed.
Through the simplification process, complex expressions become manageable, thus making calculations easier to handle and understand.