Problem 2
Question
(a) find the simplified form of the difference quotient and then (b) complete the following table. $$ \begin{array}{|c|l|l|} \hline x & h & \frac{f(x+h)-f(x)}{h} \\ \hline 5 & 2 & \\ \hline 5 & 1 & \\ \hline 5 & 0.1 & \\ \hline 5 & 0.01 & \\ \hline \end{array} $$ $$ f(x)=4 x^{2} $$
Step-by-Step Solution
Verified Answer
The simplified form is \(8x + 4h\), and the table values are 48, 44, 40.4, and 40.04.
1Step 1: Write the Expression for the Difference Quotient
The difference quotient of a function is given by \( \frac{f(x+h) - f(x)}{h} \). We need to substitute the given function \( f(x) = 4x^2 \) into this formula.
2Step 2: Substitute the Function into the Difference Quotient
Substitute \( f(x) = 4x^2 \) into the expression to get \( f(x+h) = 4(x+h)^2 \). So, the difference quotient becomes: \( \frac{4(x+h)^2 - 4x^2}{h} \).
3Step 3: Expand the Expression \(4(x+h)^2\)
Expand \( (x+h)^2 \) to get \( x^2 + 2xh + h^2 \). Thus, \( 4(x+h)^2 = 4(x^2 + 2xh + h^2) = 4x^2 + 8xh + 4h^2 \).
4Step 4: Simplify the Difference Quotient Expression
Substitute the expanded form back into the difference quotient: \( \frac{4x^2 + 8xh + 4h^2 - 4x^2}{h} \). Simplify this to \( \frac{8xh + 4h^2}{h} \).
5Step 5: Factor and Simplify Further
Factor out \( h \) from the numerator: \( \frac{h(8x + 4h)}{h} \). Cancel \( h \) from the numerator and the denominator to get \( 8x + 4h \).
6Step 6: Calculate the Values for Each \( h \) in the Table
Now use the simplified result \( 8x + 4h \) to complete the table.1. For \( x = 5, h = 2: 8(5) + 4(2) = 40 + 8 = 48 \).2. For \( x = 5, h = 1: 8(5) + 4(1) = 40 + 4 = 44 \).3. For \( x = 5, h = 0.1: 8(5) + 4(0.1) = 40 + 0.4 = 40.4 \).4. For \( x = 5, h = 0.01: 8(5) + 4(0.01) = 40 + 0.04 = 40.04 \).
Key Concepts
CalculusDerivative ApproximationFunction Analysis
Calculus
Calculus is a branch of mathematics that explores rates of change and the accumulation of quantities. It's divided primarily into two parts: differential calculus and integral calculus.
One fundamental concept in calculus is the derivative, which represents the rate of change of a function concerning a variable. This allows us to understand how a quantity increases or decreases over time or space.
Calculus uses various mathematical techniques to perform these analyses, such as limits and infinitesimals. The process of finding a derivative is called differentiation. Derivatives have practical applications in fields like physics, engineering, economics, and biology, where they are used to model real-world situations.
- Derivative finds the slope of a function at any given point. - Integrals, another part of calculus, measure the total accumulation of a quantity. Let's see how the difference quotient fits into this picture.
One fundamental concept in calculus is the derivative, which represents the rate of change of a function concerning a variable. This allows us to understand how a quantity increases or decreases over time or space.
Calculus uses various mathematical techniques to perform these analyses, such as limits and infinitesimals. The process of finding a derivative is called differentiation. Derivatives have practical applications in fields like physics, engineering, economics, and biology, where they are used to model real-world situations.
- Derivative finds the slope of a function at any given point. - Integrals, another part of calculus, measure the total accumulation of a quantity. Let's see how the difference quotient fits into this picture.
Derivative Approximation
Derivative approximation involves estimating the value of a derivative based on known values of a function.
The difference quotient is a primary tool used here. It is an expression that gives the average rate of change of a function over a small interval and is central to understanding derivatives in calculus.
The form of the difference quotient is:\[\frac{f(x+h) - f(x)}{h}\]This formula estimates the derivative of a function at a given point as \( h \) approaches zero. For the function given in the exercise, \( f(x) = 4x^2 \), the exercise uses this formula to approximate the derivative at specific intervals.- The smaller the \( h \), the closer the approximation to the actual derivative.- As seen in the exercise, as \( h \) decreases, the expression \( 8x + 4h \) provides a better approximation of the derivative.This process is essential to understand how functions behave locally.
The difference quotient is a primary tool used here. It is an expression that gives the average rate of change of a function over a small interval and is central to understanding derivatives in calculus.
The form of the difference quotient is:\[\frac{f(x+h) - f(x)}{h}\]This formula estimates the derivative of a function at a given point as \( h \) approaches zero. For the function given in the exercise, \( f(x) = 4x^2 \), the exercise uses this formula to approximate the derivative at specific intervals.- The smaller the \( h \), the closer the approximation to the actual derivative.- As seen in the exercise, as \( h \) decreases, the expression \( 8x + 4h \) provides a better approximation of the derivative.This process is essential to understand how functions behave locally.
Function Analysis
Function analysis involves studying the properties and behavior of mathematical functions.
This helps in understanding the nature of various mathematical models both theoretically and practically.
In analyzing functions, we often look for key characteristics like continuity, limits, and derivatives to better understand the behavior of a function over its domain.
This helps in understanding the nature of various mathematical models both theoretically and practically.
In analyzing functions, we often look for key characteristics like continuity, limits, and derivatives to better understand the behavior of a function over its domain.
- Continuity ensures that a function doesn't have any unexpected breaks or jumps.
- Limits help in understanding the behavior of functions at specific points or as inputs approach specific values.
- Derivatives allow us to identify how a function's values change concerning changes in its inputs.
Other exercises in this chapter
Problem 1
Differentiate two ways: first, by using the Product Rule; then, by multiplying the expressions before differentiating. Compare your results as a check. Use a gr
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Complete each of the following statements. As x approaches______________ , the value of -3x approaches 6.
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Find \(d^{2} y / d x^{2}\) $$ y=x^{5}+9 $$
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Find \(\frac{d y}{d x}\). $$ y=x^{7} $$
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