Question

Find the difference quotient of \(f\); that is, find \(\frac{f(x+h)-f(x)}{h}, h \neq 0,\) for each function. Be sure to simplify. \(f(x)=3 x^{2}-2 x+6\)

Short answer

The difference quotient is \(6x + 3h - 2.\)

Step-by-step solution

Substitute into the Difference Quotient Formula

Start by substituting the given function, \( f(x) = 3x^2 - 2x + 6 \), into the difference quotient formula: \[ \frac{f(x+h) - f(x)}{h} \]. This gives: \[ \frac{(3(x+h)^2 - 2(x+h) + 6) - (3x^2 - 2x + 6)}{h} \]

Expand and Simplify the Expression

Expand the polynomial \(3(x+h)^2 - 2(x+h) + 6 \) in the numerator: \[ 3(x+h)(x+h) - 2(x+h) + 6 = 3(x^2 + 2xh + h^2) - 2x - 2h + 6 = 3x^2 + 6xh + 3h^2 - 2x - 2h + 6. \]

Subtract the Original Function

Subtract the function \( f(x) = 3x^2 - 2x + 6 \) from the expanded form: \[ 3x^2 + 6xh + 3h^2 - 2x - 2h + 6 - (3x^2 - 2x + 6) = 3x^2 + 6xh + 3h^2 - 2x - 2h + 6 - 3x^2 + 2x - 6. \]

Simplify the Expression

Combine like terms in the numerator: \[ 3x^2 + 6xh + 3h^2 - 2x - 2h + 6 - 3x^2 + 2x - 6 = 6xh + 3h^2 - 2h. \]

Divide by h

Divide each term in the numerator by \( h \): \[ \frac{6xh + 3h^2 - 2h}{h} = \frac{6xh}{h} + \frac{3h^2}{h} - \frac{2h}{h} = 6x + 3h - 2. \]

State the Difference Quotient

The difference quotient simplified is: \[ 6x + 3h - 2. \]

Key concepts

polynomials

Polynomials are algebraic expressions that consist of variables and coefficients. They take the form of sums of monomials, such as \( ax^n \), where \( a \) is a coefficient and \( n \) is a non-negative integer exponent. In our problem, the function \( f(x) = 3x^2 - 2x + 6 \) is a second-degree polynomial. Here are some key features of polynomials:

• Degree: The highest power of the variable. In \( f(x) = 3x^2 - 2x + 6 \), the degree is 2.
• Coefficients: The numerical factors multiplying the variables. For \( f(x) \), these are 3, -2, and 6.
• Terms: Each monomial in the polynomial. The terms here are \( 3x^2 \), \( -2x \), and 6.

Polynomials can be added, subtracted, multiplied, and divided (except by zero) to form new polynomials. They are fundamental in calculus as they are easy to differentiate and integrate compared to other types of functions.

function composition

Function composition involves applying one function to the results of another. If you have two functions \( f \) and \( g \), the composition \((f \circ g)(x) \) means \( f(g(x)) \). In the context of the difference quotient, we are effectively composing \( f \) with the function that shifts \( x \) by \( h \) (i.e., \( x+h \)).

• Applying \( f \) to \( x + h \): Here, we substitute \( x + h \) into the polynomial, resulting in \( f(x+h) = 3(x+h)^2 - 2(x+h) + 6 \).
• Distributing and combining like terms: Expanding \( f(x+h) \) involves distributing and simplifying each part of the polynomial to find the new expression.

Function composition is helpful in identifying changes and shifts in functions, playing an essential role in differentiation and integration.

limits

Limits are a core concept in calculus used to define the behavior of a function as it approaches a specific point. For the difference quotient, we are interested in the limit as \( h \) approaches 0. This limit gives us the derivative of the function.

• Understanding the limit: The limit evaluates what the value of \( \frac{f(x+h) - f(x)}{h} \) approaches as \( h \) gets infinitely small. This signifies the instantaneous rate of change of the function at a point \( x \).
• Simplifying the quotient: To find the limit, we simplify the difference quotient: \( \frac{6xh + 3h^2 - 2h}{h} = 6x + 3h - 2 \).
• Taking the limit: As \( h \) approaches 0, the term \( 3h \) vanishes, simplifying our expression to \( 6x - 2 \), representing the derivative of the polynomial at any point \( x \).

Limits provide a foundational framework for calculus, underpinning the concepts of continuity, derivatives, and integrals.