Problem 62 - College Algebra Essentials

Question

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 is \(2x + h - 5\).

Step 1: Evaluate \(f(x + h)\)

Firstly, replace every \(x\) in the function \(f(x) = x^2 - 5x + 8\) with \(x + h\). This gives \(f(x + h) = (x + h)^2 - 5(x + h) + 8\). Simplify this to to get \(f(x + h) = x^2 + 2xh + h^2 - 5x - 5h + 8\).

Step 2: Substitute into the difference quotient

Substitute \(f(x + h)\) and \(f(x)\) into the difference quotient formula as follows: \(\frac{f(x + h) - f(x)}{h} = \frac{x^2 + 2xh + h^2 - 5x - 5h + 8 - (x^2 - 5x + 8)}{h}\).

Step 3: Simplify the Numerator

Next, simplify the numerator of the fraction. Notice that \(x^2\), \(-5x\) and \(8\) are cancelled by their equivalents. This simplifies the fraction to: \(\frac{2xh + h^2 - 5h}{h}\).

Step 4: Cancel the denominator

Now, factor out a \(h\) in the numerator, this allows us to cancel out the \(h\) in the denominator. The difference quotient is then simplified to: \(2x + h - 5\).

Key Concepts

College Algebra

College algebra is a foundational course that covers a variety of topics including functions, polynomials, equations, and inequalities. It serves as a stepping stone to more advanced mathematics and other subjects that require mathematical reasoning. In college algebra, you'll learn to manipulate algebraic expressions, solve equations, and understand the properties of functions.

One of the key concepts in college algebra is the difference quotient. This is a crucial concept because it forms the basis for the definition of the derivative, a central idea in calculus. Mastering the difference quotient involves being able to evaluate functions and simplify expressions—a skillset that is essential for anyone looking to succeed in college algebra and beyond.

Simplifying Expressions

To simplify mathematical expressions, especially those involving polynomials as in our textbook exercise, you need to combine like terms, factor where possible, and cancel out terms appropriately. The art of simplifying expressions is exactly that: making a complex expression into a simpler, more manageable form without changing its value.

Combining Like Terms

When simplifying the difference quotient, you’ll often encounter terms that can be combined because they have the same variable raised to the same power. For example, subtracting \(f(x)\) from \(f(x+h)\) cancels out any terms that don’t involve \(h\), making the expression simpler.

Factoring

Factoring is another important skill. In our solution, we factored out \(h\) from the numerator to cancel it with the \(h\) in the denominator. Factoring requires you to identify common factors in terms and is a powerful tool in simplification.

Function Evaluation

Function evaluation is the process of determining the output of a function given an input. In the context of algebra, it requires substituting the input value into the function and performing the arithmetic to find the result.

For example, to evaluate \(f(x+h)\) for the function \(f(x) = x^2 - 5x + 8\), we must replace \(x\) with \(x + h\) and simplify. As shown in the textbook solution, this involves not just substitution, but also expanding parentheses and combining like terms.

Mastering function evaluation enables students to work with the difference quotient and is fundamental to the understanding of more advanced mathematical concepts like limits, derivatives, and the analysis of function behavior.