Problem 40 - College Algebra

Question

In Exercises \(33-44\), find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=x^{2}-5 x+8$$

Step-by-Step Solution

Answer

The simplified difference quotient for the function \(f(x)=x^{2}-5x+8\) is \(2x + h - 5.\)

Step 1: Write down the given Function

The given function is \(f(x)=x^{2}-5x+8.\)

Step 2: Find \(f(x+h)\)

To find \(f(x+h)\), we substitute \(x+h\) into our function. Thus we get: \(f(x+h) =(x+h)^{2}-5(x+h)+8\). Simplify this to get \(f(x+h)=x^{2}+2hx+h^{2}-5x-5h+8.\)

Step 3: Substitute into the Difference Quotient

We now substitute \(f(x+h)\) and \(f(x)\) into the difference quotient: \(\frac{f(x+h)-f(x)}{h}.\) This results in: \(\frac{x^{2}+2hx+h^{2}-5x-5h+8 - (x^{2}-5x+8)}{h}\).

Step 4: Simplify the Difference Quotient

Once we simplify the difference quotient above, we get: \(\frac{2hx + h^{2} - 5h}{h}\).

Step 5: Factor out h in the numerator

When we factor out \(h\) from the numerator, we get: \(h(2x + h - 5)\).

Step 6: Final Simplification

Now, cancel out the \(h\) from the numerator and denominator and we get our simplified difference quotient: \(2x + h - 5.\)

Key Concepts

Understanding Function Simplification

Function simplification is an essential process in mathematics that makes equations easier to handle. Imagine dealing with complex equations full of terms; simplifying helps break these down into manageable pieces. In our context, we're dealing with the difference quotient, which can seem tricky at first. But don't worry, the main idea is to make it simpler and easier to work with. This process involves removing any unnecessary elements to reveal the core components. For example, when simplifying the difference quotient, we substitute the polynomial expressions and cancel out like terms wherever possible. This allows us to clearly see how the terms relate to each other without the clutter. The goal is clarity and ease, which helps us when we need to find the derivative later on.

Exploring Polynomial Functions

Polynomial functions are a fascinating and vital part of algebra. They consist of variables and coefficients and involve operations like addition, subtraction, and multiplication. Polynomials are named according to their degree: linear for first-degree, quadratic for second-degree, cubic for third-degree, and so on. In our exercise, we focus on a quadratic function:

• The function is written as: \(f(x) = x^2 - 5x + 8\).
• This tells us it has a degree of 2 (the highest power of x is 2).
• Each term holds important information: constant term \(8\), linear term \(-5x\), and quadratic term \(x^2\).

Understanding these components helps us understand the behavior of the polynomial function, such as its curvature and the nature of its graph. Recognizing how to manipulate these terms is crucial for simplifying expressions and solving equations.

Derivatives and the Role of the Difference Quotient

Derivatives are central to calculus and describe how a function changes over an interval. To think about it simply, they give us the rate of change or the slope of the function at a specific point. The difference quotient is the stepping stone to finding derivatives. It's written as \(\frac{f(x+h)-f(x)}{h}\), and it shows us how the function \(f(x)\) changes as \(x\) moves a small distance \(h\).

• Initially, you substitute \(x + h\) into the function and find the simplified expression.
• Then, you plug this back into the difference quotient, leading to an expression in terms of \(h\).
• Finally, as \(h\) approaches zero, this expression approaches the derivative of \(f(x)\).

Understanding the role of this quotient is crucial as it forms the basis for derivative calculations. By simplifying it, we can easily calculate the derivative, which helps in understanding the dynamics of a function's graph, optimizing problems, and much more in various applications.