Evaluating And Simplifying Function Expressions F(x) = 5x - 4

In this article, we will delve into the evaluation and simplification of expressions involving the function f(x) = 5x - 4. This is a fundamental concept in algebra and calculus, and understanding it thoroughly will pave the way for more advanced topics. We will explore how to find f(a), f(a + h), and the difference quotient (f(a + h) - f(a)) / h, which is a crucial concept in calculus for understanding derivatives. Each step will be explained in detail to ensure clarity and comprehension.

Evaluating f(a)

The first step in understanding function evaluation is to determine the value of f(a). When dealing with a function, evaluating f(a) simply means substituting the variable x with a in the function's expression. In our case, the function is f(x) = 5x - 4. Therefore, to find f(a), we replace every instance of x with a.

So, f(a) = 5(a) - 4. This simplifies to f(a) = 5a - 4. This expression represents the value of the function f when the input is a. It's a straightforward substitution, but it's a crucial foundation for understanding more complex evaluations. The beauty of this process lies in its simplicity; it's a direct application of the function's definition. This skill is essential for various mathematical operations, including graphing functions, solving equations, and understanding transformations.

To further illustrate this concept, let's consider a few examples. If a = 2, then f(2) = 5(2) - 4 = 10 - 4 = 6. Similarly, if a = -1, then f(-1) = 5(-1) - 4 = -5 - 4 = -9. These examples highlight how the value of f(a) changes depending on the value of a. Understanding this relationship between input and output is fundamental to grasping the behavior of functions. Moreover, this evaluation process is not limited to simple numerical values; a can represent any algebraic expression, further expanding the applicability of this concept. The ability to correctly evaluate functions for various inputs is a cornerstone of mathematical proficiency.

Determining f(a + h)

Now, let's move on to evaluating f(a + h). This is a slightly more complex evaluation than f(a), but the underlying principle remains the same: we substitute the variable x in the function with the expression (a + h). Given our function f(x) = 5x - 4, we replace x with (a + h) to get f(a + h) = 5(a + h) - 4.

The next step is to simplify this expression. We distribute the 5 across the terms inside the parentheses: 5(a + h) = 5a + 5h. So, f(a + h) = 5a + 5h - 4. This is the simplified form of f(a + h). It's important to note that we cannot combine the terms 5a, 5h, and -4 because they are not like terms. 5a and 5h contain variables, while -4 is a constant. This expression is crucial in calculus, particularly when dealing with the concept of the difference quotient and derivatives.

Understanding how to correctly evaluate f(a + h) is paramount for grasping the idea of incremental change in a function. The h in (a + h) represents a small change in the input, and f(a + h) gives us the function's value at this slightly altered input. This is a fundamental concept in understanding rates of change and slopes of curves. To solidify this concept, consider if a = 1 and h = 0.5, then f(1 + 0.5) = f(1.5) = 5(1.5) - 4 = 7.5 - 4 = 3.5. This numerical example provides a concrete understanding of how f(a + h) represents the function's output when the input is slightly perturbed. The ability to manipulate and simplify such expressions is a vital skill in both algebra and calculus.

Simplifying the Difference Quotient: (f(a + h) - f(a)) / h

The most complex expression we will tackle is the difference quotient, (f(a + h) - f(a)) / h. This expression is fundamental in calculus as it represents the average rate of change of the function f over a small interval of length h. It's the foundation for understanding the derivative, which gives us the instantaneous rate of change.

We already know that f(a + h) = 5a + 5h - 4 and f(a) = 5a - 4. Now, we substitute these expressions into the difference quotient:

[(5a + 5h - 4) - (5a - 4)] / h

The next step is to simplify the numerator. We distribute the negative sign in the second term:

5a + 5h - 4 - 5a + 4

Notice that the 5a and -5a terms cancel each other out, and the -4 and +4 terms also cancel each other out. This leaves us with:

5h

So, the difference quotient becomes:

(5h) / h

Now, we can simplify by dividing both the numerator and the denominator by h (assuming h is not zero): (5h) / h = 5.

Therefore, the simplified difference quotient is 5. This result is particularly significant because it represents the slope of the line represented by the function f(x) = 5x - 4. In general, the difference quotient provides a way to calculate the slope of a function at a specific point, which is a core concept in differential calculus. This exercise demonstrates the power of algebraic manipulation in revealing fundamental properties of functions. Understanding the difference quotient is a critical step in transitioning from basic algebra to the more advanced concepts of calculus. This simplification process highlights the elegance and efficiency of mathematical tools in uncovering underlying relationships and patterns.

Conclusion

In summary, we have successfully evaluated and simplified the expressions f(a), f(a + h), and the difference quotient (f(a + h) - f(a)) / h for the function f(x) = 5x - 4. These operations are fundamental to understanding functions and their behavior. Evaluating f(a) is a straightforward substitution, while evaluating f(a + h) involves distribution and simplification. The difference quotient, though more complex, is a crucial concept in calculus for understanding rates of change and derivatives.

The ability to manipulate and simplify these expressions is a valuable skill in mathematics. It allows us to analyze functions, solve equations, and understand the relationships between variables. The difference quotient, in particular, serves as a bridge between algebra and calculus, paving the way for more advanced mathematical concepts. By mastering these fundamental techniques, you will be well-equipped to tackle more challenging problems in mathematics and related fields. The journey through these evaluations and simplifications not only reinforces algebraic skills but also provides a glimpse into the power and elegance of mathematical reasoning.