Problem 47 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 47-48: Simplify the difference quotient $$ \frac{f(2+h)-f(2)}{h} \text { if } h \neq 0 . $$ $$ f(x)=x^{2}-3 x $$

Step-by-Step Solution

Answer

h + 1.

Step 1: Substitute into the Function

First, we need to substitute \(x = 2 + h\) and \(x = 2\) into the function \(f(x) = x^2 - 3x\).Calculate \(f(2 + h)\):\[f(2+h) = (2+h)^2 - 3(2+h)\]Expand the square:\[(2+h)^2 = 4 + 4h + h^2\]Therefore,\[f(2+h) = 4 + 4h + h^2 - 6 - 3h = h^2 + h - 2\]Calculate \(f(2)\):\[f(2) = 2^2 - 3 \times 2 = 4 - 6 = -2\]

Step 2: Construct the Difference Quotient

Now that we have \(f(2+h) = h^2 + h - 2\) and \(f(2) = -2\), substitute these into the difference quotient:The difference quotient is:\[\frac{f(2+h) - f(2)}{h} = \frac{(h^2 + h - 2) - (-2)}{h}\]Simplify the expression:\[\frac{h^2 + h - 2 + 2}{h} = \frac{h^2 + h}{h}\]

Step 3: Simplify the Quotient

Simplify \(\frac{h^2 + h}{h}\). Each term in the numerator contains \(h\), so factor out \(h\):\[\frac{h(h + 1)}{h}\]Cancel \(h\) from the numerator and the denominator:\[h + 1\]Since \(h eq 0\), the simplified form of the difference quotient is \(h + 1\).

Key Concepts

Polynomial Function

A polynomial function is a mathematical expression dealing with terms that are made up of constants, variables, and exponents. For students handling calculus and algebra, it is crucial to understand these basic components of a polynomial function because they appear frequently.Polynomials are usually expressed in the form:\[a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\]Where:

• \(a_n, a_{n-1}, \ldots, a_0\) are coefficients
• \(x\) is the variable
• \(n\) is a non-negative integer

In our exercise, the polynomial given is: \[f(x) = x^2 - 3x\]Here, we have a quadratic polynomial because the highest power of \(x\) is 2. Studying polynomial functions prepares students to engage with key mathematical operations, such as differentiation, which is the process of finding a derivative.

Function Substitution

Function substitution involves replacing the variable \(x\) in a function with another value or expression. This is a critical step when you're working with functions, especially in differentiating or finding function values over intervals.For the exercise at hand, we aim to substitute \((2 + h)\) and \(2\) into the polynomial function \(f(x) = x^2 - 3x\). This process can be detailed as follows:- **Substitute \(2 + h\):** To find \(f(2 + h)\), replace \(x\) with \(2 + h\): \[ f(2+h) = (2+h)^2 - 3(2+h) \]- **Substitute \(2\):** To find \(f(2)\), replace \(x\) with \(2\): \[ f(2) = 2^2 - 3 imes 2 \]By applying substitution, you explore various values within a function to analyze behavior and characteristics.

Simplification

The simplification process in mathematics allows us to rewrite an equation or expression in a more understandable or practical form. This often makes problem-solving much easier.During the exercise, simplification was applied after substituting the expressions in the function. Here are some steps involved:- **Expanding Squares:** \[(2+h)^2 = 4 + 4h + h^2\] This expansion helps in eliminating parentheses.- **Simplifying the Expression:** Combine like terms. For instance, \[4 + 4h + h^2 - 6 - 3h = h^2 + h - 2\] Remove unnecessary constants and combine similar terms.The resulting simpler expression allows for further solving or for finding derivatives, which is often the next step in calculus.

Factoring

Factoring is breaking down an expression into simpler terms or factors that can be multiplied together to obtain the original expression. It is a necessary skill that has numerous applications in algebra.In the solution, once the polynomial was substituted and simplified, factoring was needed to make further simplification:- **Expression to be factored:** \[h^2 + h\]- **Factoring out \(h\):** Both terms in the expression contain \(h\), which allows us to factor \(h\): \[h(h + 1)\]- **Canceling \(h\):** When placed over \(h\) in the expression \[\frac{h(h + 1)}{h}\] Cancel \(h\) (not including division by zero cases): \[h + 1\]Factoring is an essential concept that simplifies expressions, making them easier to manipulate, which is particularly handy when finding the limit of a function.