Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

, find \(\lim _{x \rightarrow 2}[f(x)-f(2)] /(x-2)\) for each given function \(\bar{f}\) $$ f(x)=3 x^{2}+2 x+1 $$

Step-by-Step Solution

Answer

The limit is 14.

Step 1: Substitute Directly to Verify Setup

First, evaluate the function at the point where the limit approaches, which is at \(x = 2\). Substitute \(x = 2\) into \(f(x) = 3x^{2} + 2x + 1\) to find \(f(2)\). Calculating: \(f(2) = 3(2)^{2} + 2(2) + 1 = 12 + 4 + 1 = 17\).

Step 2: Apply the Difference Formula

Next, substitute \(f(x)\) and \(f(2)\) into the limit form, \(\lim_{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}\), which becomes \(\lim_{x \rightarrow 2} \frac{3x^2 + 2x + 1 - 17}{x - 2}\). Simplify the expression to \(\lim_{x \rightarrow 2} \frac{3x^2 + 2x - 16}{x - 2}\).

Step 3: Factor the Numerator

To simplify the expression, factor the numerator, \(3x^2 + 2x - 16\). This can be factored into \((3x - 4)(x + 4)\), so the expression becomes \(\lim_{x \rightarrow 2} \frac{(3x - 4)(x + 4)}{x - 2}\).

Step 4: Cancel the Common Terms

Notice that \(x = 2\) does not appear to directly cancel any terms in the factorization from Step 3. Therefore, re-evaluating the factorization is necessary. Upon checking, the expression \(3x^2 + 2x - 16\) incorrectly factors over integers for simple roots, hence at Step 3 it should be managed with direct substitution approach if failing factor simplification.

Step 5: Evaluate the Limit by Substitution

Evaluate the derivative of \(f(x)\) to resolve the issue correctly, finding \(f'(x)\). The derivative \(f'(x) = 6x + 2\) represents the slope as traditional limit approach resolves to the derivative interpretation. Evaluating it at \(x=2\), \(f'(2) = 6(2) + 2 = 14\). So, \(\lim _{x \rightarrow 2}\frac{f(x)-f(2)}{x-2} = 14\).

Key Concepts

Difference Quotient

The concept of the **difference quotient** is foundational in calculus, serving as a stepping stone to understanding derivatives. It's essentially a way to determine how a function changes at a particular point.

• The difference quotient is typically expressed as:\[\frac{f(x) - f(a)}{x - a}\]where \( f(x) \) is the function in question, and the difference \( x-a \) represents the distance between two points on a curve.
• As the difference \( x - a \) approaches zero, the difference quotient approaches the derivative at \( x = a \). This process helps measure the instantaneous rate of change, or slope, at a single point.

Understanding the difference quotient requires careful attention to simplification, especially in polynomial expressions, to ensure that no factors are extraneously ignored.
This careful simplification sets up the stepping stone for evaluating limits effectively.

Polynomial Function

**Polynomial functions** are versatile and commonly encountered in calculus. They are named for the powers, or 'degrees', of their variables.

• A polynomial function can be written as a sum of terms, consisting of variables raised to whole-number exponents, multiplied by coefficients. For example, \( f(x) = 3x^2 + 2x + 1 \) is a polynomial of degree 2 (quadratic).
• The degree of a polynomial is significant because it influences the function's shape and the number of roots it may have.

When working with polynomial functions in calculus, operations like addition, subtraction, and multiplication remain straightforward. However, dividing polynomials, often needed when dealing with the difference quotient, can demand greater attention to algebraic manipulation and factorization, ensuring that limits can be evaluated accurately and smoothly.

Derivative

The **derivative** provides a way to calculate the rate at which a function is changing at any given point. It's a core concept in calculus.

• In the context of a polynomial function, the derivative can be derived using standard rules, such as the power rule. For example, for the polynomial \( f(x) = 3x^2 + 2x + 1 \), the derivative \( f'(x) \) is calculated as:\[ 6x + 2 \]
• This formula reveals how fast \( f(x) \) changes with respect to \( x \), representing the slope of the tangent line to the curve of \( f \) at any point \( x \).
• Derivatives are crucial within limit evaluation, as recognizing the result of a difference quotient as a derivative simplifies the computation and offers insights into a function's behavior around that point, such as concavity and inflection.

Limit Evaluation

**Limit evaluation** is the process of finding the value a function approaches as the input approaches a certain point. It's a vital concept for understanding behavior in calculus.

• Limits handle points of potential division by zero or discontinuity. For example, evaluating \[\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}\]requires simplifying \( \frac{f(x)-f(2)}{x-2} \) so as to eliminate the \( x-2 \) term and avoid division by zero.
• Here, the limit manifests the derivative's evaluation at a particular \( x \, (x = 2) \). Therefore, understanding the limit transforms this solution into the calculation of \( f'(2) \), providing 14 as the solution.
• Through mastering limit evaluation, you gain the ability to resolve indeterminate forms and comprehend the lasting implications of a function's behavior at crucial points like roots or discontinuities.

Limit evaluation not only bolsters understanding but also reinforces the systematic nature of calculus, showcasing the elegant transitions between algebra and calculus.