Problem 49 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 49-50: Simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h} \text { if } h \neq 0 . $$ $$ f(x)=x^{2}+5 $$

Step-by-Step Solution

Answer

The simplified difference quotient is \( 2x + h \).

Step 1: Substitute f(x) and f(x+h) into the Difference Quotient

First, identify the function \( f(x) = x^2 + 5 \). Use this function to find \( f(x+h) \), which means replacing \( x \) with \( x + h \). So, \( f(x+h) = (x+h)^2 + 5 \).

Step 2: Expand and Simplify f(x+h)

Expand \( (x+h)^2 \) to get \( x^2 + 2xh + h^2 \). Substitute this into \( f(x+h) \) to get \( f(x+h) = x^2 + 2xh + h^2 + 5 \).

Step 3: Compute f(x+h) - f(x)

Find the difference \( f(x+h) - f(x) = (x^2 + 2xh + h^2 + 5) - (x^2 + 5) \). Simplify to \( 2xh + h^2 \).

Step 4: Substitute into the Difference Quotient

Now substitute the result from Step 3 into the difference quotient \( \frac{f(x+h) - f(x)}{h} \). This gives: \( \frac{2xh + h^2}{h} \).

Step 5: Simplify the Expression

Factor \( h \) out of the numerator to get \( h(2x + h) \). Then, simplify the fraction \( \frac{h(2x + h)}{h} \) by canceling \( h \), resulting in \( 2x + h \).

Key Concepts

Simplifying Algebraic Expressions

Simplifying algebraic expressions is about making them easier to work with by reducing their complexity. This usually involves eliminating unnecessary terms and factoring out common factors. The key is to express the expression in its simplest form.
Consider an example like the difference quotient from the exercise:

• First, expand expressions like \( (x+h)^2 \) to \( x^2 + 2xh + h^2 \).
• Next, eliminate like terms or factors to reduce complexity. In our exercise, we subtract \( f(x) \) from \( f(x+h) \), yielding \( 2xh + h^2 \).

By dividing by \( h \) and canceling it out, the expression \( 2x + h \) is more straightforward. This simplification is useful for understanding functions' behaviors as inputs change.

Function Notation

Function notation is a way of representing functions mathematically. It is designed to clarify which variable is the input and what operation is being performed.
In the exercise, \( f(x) = x^2 + 5 \) is the function notation where:

• \( f \) indicates we are dealing with a function.
• \( x \) is the input variable.
• \( x^2 + 5 \) is the rule applied to the input.

To find \( f(x+h) \), the input changes to \( x+h \). This shows precisely how each element of the function adjusts with the change, which is crucial for calculations involving rates of change.
Understanding function notation helps you manipulate and interpret mathematical models involving functions easily.

Polynomial Functions

Polynomial functions are mathematical expressions that consist of variables raised to whole number powers and their corresponding coefficients. These functions can take on many forms but are vital for modeling many real-world phenomena.
In the provided exercise, \( f(x) = x^2 + 5 \) is a polynomial function. Here:

• The highest degree is 2, making it a quadratic polynomial.
• It includes constants like 5, which does not change with \( x \).
• The form \( ax^2 + bx + c \) can describe it, with \( a = 1 \), \( b = 0 \), and \( c = 5 \).

Polynomial functions are flexible and powerful tools in math. They help solve problems and describe situations like parabolic paths, optimization, and natural shapes. Understanding how to work with them, such as moving from one form to another, simplifies complex problems.