Problem 155 - Precalculus Essentials

Question

If \(f(x)=x^{2}+3 x+2,\) find \(\frac{f(x+h)-f(x)}{h}, h \neq 0,\) and simplify.

Step-by-Step Solution

Answer

Therefore, \(\frac{f(x+h)-f(x)}{h}= 2x + h + 3\).

Step 1: Substitute \(x+h\) into \(f(x)\)

First, substitute \(x+h\) into the function \(f(x)\), this gives \(f(x+h) = (x+h)^{2}+3(x+h)+2\). Expand this to give \(f(x+h) = x^{2}+2xh+h^{2}+3x+3h+2\).

Step 2: Compute \(f(x+h)-f(x)\)

Next, compute \(f(x+h)-f(x)\). Since \(f(x)=x^{2}+3x+2\), therefore \(f(x+h)-f(x) = (x^{2}+2xh+h^{2}+3x+3h+2) - (x^{2}+3x+2)\). Simplify this to be \(2xh + h^{2} + 3h\).

Step 3: Divide by \(h\)

Now, divide by \(h\) to get \(\frac{f(x+h)-f(x)}{h} = \frac{2xh + h^{2} + 3h}{h}\). Simplify this to give \(2x + h + 3\).

Key Concepts

Polynomial Function

A polynomial function is a mathematical expression comprising variables and coefficients. In simpler terms, it's an equation that involves powers of a variable, commonly represented as 'x'. For example, in the function given in the exercise, \(f(x) = x^2 + 3x + 2\), the variable is \(x\), and the coefficients are numbers like 1 (for \(x^2\)), 3, and 2.
Polynomial functions are foundational in algebra because they are versatile and expressive. They're used to model various phenomena, like quadratic motion or growth patterns, thanks to their ability to accommodate multiple degrees through the powers of \(x\).

• Linear functions (polynomials of degree 1) have the form \(ax + b\).
• Quadratic functions (polynomials of degree 2) include terms like \(ax^2 + bx + c\).
• Higher-degree polynomial functions involve terms like \(ax^n\) where \(n\) is any whole number greater than 2.

Understanding polynomial functions is integral to tackling more advanced mathematical concepts, such as calculus, which frequently involves blending polynomial functions with limits and derivatives.

Simplification

Simplification in algebra means reducing expressions to their most basic or concise forms. The goal is to make the expression easier to work with, understand, or compute. In the original exercise, the expression \(f(x+h) - f(x)\) eventually gets simplified to \(2x + h + 3\).
The process involved several steps:

• Expanding expressions like \((x+h)^2\) to \(x^2 + 2xh + h^2\).
• Subtracting identical terms to eliminate parts where variables cancel out.
• Breaking down terms and finding common factors when necessary.

Simplification is crucial, not just for neatness but also to make calculations feasible and to set up further algebraic manipulations such as solving equations or evaluating limits.

Derivative Concepts

The derivative is a core concept in calculus that represents the rate at which a function is changing at any given point. The difference quotient, as shown in the exercise, \(\frac{f(x+h)-f(x)}{h}\), signifies an initial step in finding a derivative. This approach estimates how \(f(x)\) changes as \(x\) shifts by a small increment \(h\).
Understanding this is essential for conceptually grasping physical phenomena like velocity and acceleration.

• The difference quotient defines the slope of the secant line connecting two points on a function's graph. As \(h\) approaches zero, this slope becomes the derivative, or rate of change, at a specific point.
• This exercise simplifies the difference quotient to \(2x + h + 3\), showing the line's slope connecting points increasingly close together.
• In calculus, derivatives provide tools to determine maxima, minima, and inflection points, offering insights into the behavior of functions.

By practicing such exercises, students build a foundation for tackling real-world problems through mathematical modeling and analysis using derivatives.