Problem 87 - A Graphical Approach to College Algebra

Question

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) ) for each function \(f\). Simplify completely. $$f(x)=3 x^{2}$$

Step-by-Step Solution

Answer

The difference quotient simplifies to \(6x + 3h\).

Step 1: Set Up the Difference Quotient

The difference quotient is calculated as \(\frac{f(x+h)-f(x)}{h}\), where \(f(x) = 3x^2\). Start by calculating \(f(x+h)\) by substituting \(x+h\) into the function. This gives \(f(x+h) = 3(x+h)^2\).

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

Expand \((x+h)^2\) to get \(x^2 + 2xh + h^2\). Substitute this back into \(f(x+h)\), yielding \(f(x+h) = 3(x^2 + 2xh + h^2) = 3x^2 + 6xh + 3h^2\).

Step 3: Substitute into the Difference Quotient Formula

Now, substitute \(f(x+h) = 3x^2 + 6xh + 3h^2\) and \(f(x) = 3x^2\) into the difference quotient: \[\frac{3x^2 + 6xh + 3h^2 - 3x^2}{h}\].

Step 4: Simplify the Numerator

Subtract \(3x^2\) from \(3x^2 + 6xh + 3h^2\). The \(3x^2\) terms cancel out, leaving \(6xh + 3h^2\). Thus, the expression becomes \(\frac{6xh + 3h^2}{h}\).

Step 5: Simplify the Expression

Factor \(h\) out of the numerator: \(6xh + 3h^2 = h(6x + 3h)\), allowing it to cancel with \(h\) in the denominator: \[\frac{h(6x + 3h)}{h} = 6x + 3h\].

Step 6: Present the Simplified Result

The simplified difference quotient after cancellation is \(6x + 3h\).

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables, where each term consists of a variable raised to a non-negative integer power and multiplied by a coefficient. For example, in the function given in the original exercise, \(f(x) = 3x^2\), we have a simple polynomial:

• 3 is the coefficient of the term.
• x^2 indicates the degree of the polynomial, which is 2 in this case, meaning it's a quadratic polynomial function.

Polynomial functions can have different degrees, such as linear (degree 1), quadratic (degree 2), cubic (degree 3), etc. Understanding these types of functions and their components is essential in algebra, as they form the basis for many other mathematical concepts.
When dealing with polynomial functions, it's crucial to identify each term and understand how to manipulate these components, which is often required when performing tasks like differentiation using difference quotients.

Simplification

Simplification in mathematics is the process of transforming an expression into its most reduced form, making it easier to work with or understand. There are several reasons we simplify expressions:

• To make calculations easier by reducing the number of terms.
• To reveal the underlying relationships or structures within the expression.
• To make it easier to identify key features, such as intercepts or asymptotes in functions.

In the example of simplifying the difference quotient \(\frac{3x^2 + 6xh + 3h^2 - 3x^2}{h}\), we can see simplification in action. By canceling out terms and factoring, we are left with \(6x + 3h\). This reduction not only makes the quotient easier to interpret, but it also provides insight into how the function behaves as \(h\) approaches zero.

Algebraic Manipulation

Algebraic manipulation involves using mathematical operations and properties to change the form of an expression or equation. This can include operations such as expansion, factoring, combining like terms, and simplification. In our example of the difference quotient, algebraic manipulation is necessary to break down and reconfigure the function expression:

• Expanding \((x+h)^2\) involves recognizing the algebraic identity \((a+b)^2 = a^2 + 2ab + b^2\), leading to \(x^2 + 2xh + h^2\).
• When substituting back into the function, re-distribute coefficients, like 3 over every term.
• Cancel out terms in the difference quotient to simplify it.

Algebraic manipulation is crucial for deriving results such as limits and derivatives, as it allows us to precisely handle expressions to arrive at meaningful results. This process helps us to refine our understanding of the behavior of functions and solve equations efficiently.