Problem 108
Question
In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=3 x^{2}+2 x$$
Step-by-Step Solution
Verified Answer
The difference quotient of the function \(f(x)=3x^{2}+2x\) is \(6x+3h+2\).
1Step 1: Insert the function into the difference quotient
First, substitute the function \(f(x) = 3x^{2}+2x\) into the expression of the difference quotient. This results in: \(\frac{f(x+h)-f(x)}{h}= \frac{(3(x+h)^{2}+2(x+h))-(3x^{2}+2x)}{h}\)
2Step 2: Simplify the numerator
The next step is to simplify the numerator. Expand and simplify the terms to get: \(\frac{3x^{2}+6xh+3h^{2}+2x+2h-3x^{2}-2x}{h}= \frac{6xh+3h^{2}+2h}{h}\)
3Step 3: Divide Each Term by \(h\)
To further simplify the expression, divide each term in the numerator by \(h\). Therefore, the difference quotient becomes: \(6x+3h+2\)
Key Concepts
PrecalculusAlgebraic FunctionsDerivatives
Precalculus
Precalculus is a foundational area of mathematics that prepares students for calculus. It combines the skills and concepts from algebra and trigonometry to tackle more advanced mathematical problems.
A crucial topic covered in precalculus is the difference quotient, which provides a way to approximate the slope of the tangent line to a curve at a point. This concept becomes essential in understanding derivatives, a core component of calculus.
The difference quotient formula, \(\frac{f(x+h)-f(x)}{h}\), is used as an introduction to the idea of a derivative. By working with this formula, students start to see how small changes in \(x\) affect the function \(f(x)\), leading to a deeper understanding of instantaneous rates of change.
Precalculus lays the groundwork by building on existing algebraic reasoning skills, introducing new types of functions, and encouraging problem-solving strategies that are vital for tackling calculus.
A crucial topic covered in precalculus is the difference quotient, which provides a way to approximate the slope of the tangent line to a curve at a point. This concept becomes essential in understanding derivatives, a core component of calculus.
The difference quotient formula, \(\frac{f(x+h)-f(x)}{h}\), is used as an introduction to the idea of a derivative. By working with this formula, students start to see how small changes in \(x\) affect the function \(f(x)\), leading to a deeper understanding of instantaneous rates of change.
Precalculus lays the groundwork by building on existing algebraic reasoning skills, introducing new types of functions, and encouraging problem-solving strategies that are vital for tackling calculus.
Algebraic Functions
Algebraic functions are expressions that involve constants and variables, using operations like addition, subtraction, multiplication, division, and exponentiation involving whole numbers.
In the given exercise, we deal with the algebraic function \(f(x) = 3x^2 + 2x\). This quadratic function consists of terms raised to the power of 2 and 1, making it a polynomial.
In the context of the difference quotient, students learn how to manipulate these expressions by expanding polynomials and simplifying terms.
In the given exercise, we deal with the algebraic function \(f(x) = 3x^2 + 2x\). This quadratic function consists of terms raised to the power of 2 and 1, making it a polynomial.
In the context of the difference quotient, students learn how to manipulate these expressions by expanding polynomials and simplifying terms.
- Substitution: Inserting \(x + h\) into the function, as shown in \((3(x+h)^2 + 2(x+h))\).
- Expansion: Expanding the squared term \((x+h)^2\) and distributing the coefficients.
- Simplification: Collecting like terms and reducing the expression.
Derivatives
Derivatives are one of the most significant concepts in calculus, describing how a function changes as its input changes.
The difference quotient \(\frac{f(x+h)-f(x)}{h}\), is a fundamental building block for understanding derivatives. It approximates the rate of change of a function over a small interval \(h\).
When \(h\) approaches zero, the difference quotient becomes the derivative of \(f(x)\). In our example, simplifying the expression yields \(6x + 3h + 2\). As \(h\) goes to zero, the derivative is \(6x + 2\).
This result indicates how steeply \(f(x) = 3x^2 + 2x\) increases at any point \(x\). The derivative function helps identify critical points, such as maxima, minima, and points of inflection. Understanding derivatives allows students to not just analyze change, but also predict behavior, and optimize various problems in calculus.
The difference quotient \(\frac{f(x+h)-f(x)}{h}\), is a fundamental building block for understanding derivatives. It approximates the rate of change of a function over a small interval \(h\).
When \(h\) approaches zero, the difference quotient becomes the derivative of \(f(x)\). In our example, simplifying the expression yields \(6x + 3h + 2\). As \(h\) goes to zero, the derivative is \(6x + 2\).
This result indicates how steeply \(f(x) = 3x^2 + 2x\) increases at any point \(x\). The derivative function helps identify critical points, such as maxima, minima, and points of inflection. Understanding derivatives allows students to not just analyze change, but also predict behavior, and optimize various problems in calculus.
Other exercises in this chapter
Problem 106
In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=-2 x+3$$
View solution Problem 107
In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=-x^{2}+x$$
View solution Problem 109
In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=\frac{1}{x-3}, x \neq 3$$
View solution Problem 110
In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=\frac{1}{x+1}, x \neq-1$$
View solution