/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 The table shows values of \(f(x)... [FREE SOLUTION] | 91Ó°ÊÓ

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The table shows values of \(f(x)=x^{3}\) near \(x=2\) (to three decimal places). Use it to estimate \(f^{\prime}(2)\) $$\begin{array}{c|ccccc}\hline x & 1.998 & 1.999 & 2.000 & 2.001 & 2.002 \\\\\hline x^{3} & 7.976 & 7.988 & 8.000 & 8.012 & 8.024 \\\\\hline\end{array}$$

Short Answer

Expert verified
The approximate derivative \( f'(2) \) is 12.

Step by step solution

01

Understand the Problem

We need to estimate the derivative of the function \( f(x) = x^3 \) at \( x = 2 \), which is denoted by \( f'(2) \). The derivative \( f'(x) \) represents the slope of the tangent line to the function at any given point \( x \).
02

Recall the Definition of Derivative

The derivative \( f'(x) \) can be approximated using the difference quotient: \[ f'(x) \approx \frac{f(x + h) - f(x)}{h} \] where \( h \) is a small number. We'll apply this formula to approximate \( f'(2) \).
03

Choose Values for Approximation

To estimate \( f'(2) \), we'll use the points around \( x = 2 \). Specifically, we can use \( x = 2.001 \) and \( x = 1.999 \) to find the average rate of change around \( x = 2 \).
04

Calculate Approximations Using Forward Difference

Using the forward difference with data point \( x = 2.001 \):\[ f'(2) \approx \frac{f(2.001) - f(2)}{2.001 - 2} = \frac{8.012 - 8.000}{0.001} = 12 \]
05

Calculate Approximations Using Backward Difference

Using the backward difference with data point \( x = 1.999 \):\[ f'(2) \approx \frac{f(2) - f(1.999)}{2 - 1.999} = \frac{8.000 - 7.988}{0.001} = 12 \]
06

Average the Two Approximations

Average the forward and backward approximations to get a more accurate estimate:\[ f'(2) = \frac{12 + 12}{2} = 12 \]
07

Conclusion

Based on the calculations, the approximate value of \( f'(2) \) is 12, which represents the slope of the tangent line to \( f(x) \) at \( x = 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference Quotient
The difference quotient is a fundamental concept in calculus used to approximate the derivative of a function. In simpler terms, it is a formula that provides a way to compute the slope of a function over an interval. This slope is calculated between two points on the function, and it is given by \[ f'(x) \approx \frac{f(x + h) - f(x)}{h} \] where \( h \) represents a small step or interval between the two points.
The idea is to use points that are very close together to get a slope that almost matches the exact slope at a particular point.
This is helpful because it allows us to find the rate at which a function is changing at any given moment.
  • By selecting values of \( x \) that are near the point of interest, we can closely estimate this slope.
  • The smaller the value of \( h \), the more accurate the approximation tends to be.
Slope of Tangent Line
The slope of a tangent line to a curve at a given point can be understood as the instantaneous rate of change at that point. For a function such as \( f(x) = x^3 \), the tangent line's slope at \( x = 2 \) corresponds to the derivative \( f'(2) \).
In visualization terms, if you draw a line that just "touches" the curve at one exact point without crossing it, that line is the tangent.
Its slope tells us how steep or flat the curve is at that specific point:
  • A steep slope implies a rapid change in the function value.
  • A flat slope denotes a slower change.
The tangent line slope is what makes calculus so powerful, giving precise insights into a function's behavior exactly at a point rather than over an interval.
Average Rate of Change
The average rate of change is a straightforward concept that essentially tells us how much a function's value is changing, on average, over a specific interval. In mathematical terms, it is given by the same formula as the difference quotient but interpreted over a larger interval.
Using our previous example with function values near \( x = 2 \), the average rate of change involves calculating how much \( f(x) = x^3 \) changes as \( x \) moves from one value, say 1.999, to another, like 2.001.
By looking at this rate over small intervals around a point, we can use these values to estimate the derivative.
  • This helps in finding the instantaneous rate of change, which is essentially the derivative, by averaging values over very small intervals.
  • The smaller the interval around a point, the better the estimate of the derivative becomes.
Ultimately, finding the average rate of change over successively smaller intervals helps us inch closer to the precise slope of the tangent line at any specific point.

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Most popular questions from this chapter

Water is flowing into a tank; the depth, in feet, of the water at time \(t\) in hours is \(h(t) .\) Interpret, with units, the following statements. (a) \(\quad h(5)=3\) (b) \(h^{\prime}(5)=0.7\) (c) \(h^{-1}(5)=7\) (d) \(\quad\left(h^{-1}\right)^{\prime}(5)=1.2\)

Are the statements true or false? Give an explanation for your answer. If \(f^{\prime}(x)\) is increasing, then \(f(x)\) is also increasing.

Find the derivatives algebraically. $$f(x)=5 x^{2}\( at \)x=10$$

A laboratory study investigating the relationship between diet and weight in adult humans found that the weight of a subject, \(W,\) in pounds, was a function, \(W=f(c),\) of the average number of Calories per day, \(c,\) consumed by the subject. (a) In terms of diet and weight, interpret the statements $$f(1800)=155, f^{\prime}(2000)=0, \text { and } f^{-1}(162)= 2200.$$ (b) What are the units of \(f^{\prime}(c)=d W / d c ?\)

When you breathe, a muscle (called the diaphragm) reduces the pressure around your lungs and they expand to fill with air. The table shows the volume of a lung as a function of the reduction in pressure from the diaphragm. Pulmonologists (lung doctors) define the compliance of the lung as the derivative of this function. \(^{9}\) (a) What are the units of compliance? (b) Estimate the maximum compliance of the lung. (c) Explain why the compliance gets small when the lung is nearly full (around 1 liter). $$\begin{array}{|c|c|}\hline \begin{array}{c}\text { Pressure reduction } \\\\\text { (cm of water) } \end{array} & \begin{array}{c}\text { Volume } \\\\\text { (liters) }\end{array} \\\\\hline 0 & 0.20 \\ \hline 5 & 0.29 \\\\\hline 10 & 0.49 \\\\\hline 15 & 0.70 \\\\\hline 20 & 0.86 \\\\\hline 25 & 0.95 \\ \hline 30 & 1.00 \\\\\hline\end{array}$$

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