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Chapter 3: Problem 30

Each function \(f(x)\) changes value when \(x\) changes from \(x_{0}\) to \(x_{0}+d x .\) Find a. the change \(\Delta f=f\left(x_{0}+d x\right)-f\left(x_{0}\right)\) b. the value of the estimate \(d f=f^{\prime}\left(x_{0}\right) d x ;\) and c. the approximation error \(|\Delta f-d f|\) \(f(x)=2 x^{2}+4 x-3, \quad x_{0}=-1, \quad d x=0.1\)

Short Answer

Expert verified
\( \Delta f = 0.02 \), \( df = 0 \), error = 0.02.

Step by step solution

01

Calculate \( \Delta f \)

To find the change \( \Delta f \), compute \( f(x_0 + dx) \) and \( f(x_0) \). We start by calculating \( f(x_0+dx) = f(-1 + 0.1) = f(-0.9) \). Substitute \( x = -0.9 \) into the function: \[ f(-0.9) = 2(-0.9)^2 + 4(-0.9) - 3 \] which simplifies to \[ f(-0.9) = 2(0.81) - 3.6 - 3 = 1.62 - 3.6 - 3 = -4.98 \]. Now, calculate \( f(x_0) = f(-1) \): \[ f(-1) = 2(-1)^2 + 4(-1) - 3 = 2(1) - 4 - 3 = 2 - 4 - 3 = -5 \]. Therefore, \( \Delta f = f(-0.9) - f(-1) = -4.98 - (-5) = 0.02 \).
02

Calculate \( df \) using Derivative

To find the estimate \( df \), first determine the derivative of \( f(x) = 2x^2 + 4x - 3 \), which is \( f'(x) = 4x + 4 \). Plug in \( x_0 = -1 \) to get \( f'(-1) = 4(-1) + 4 = -4 + 4 = 0 \). Then, compute \( df = f'(-1) \cdot dx = 0 \cdot 0.1 = 0 \).
03

Calculate Approximation Error \( |\Delta f - df| \)

Compute the absolute difference between \( \Delta f \) and \( df \): \( |\Delta f - df| = |0.02 - 0| = 0.02 \).
04

Conclusion

We have calculated all parts of the exercise. \( \Delta f = 0.02 \), \( df = 0 \), and the approximation error is 0.02.

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Key Concepts

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

Function Changes
When dealing with functions, understanding how they change as inputs vary is crucial. Consider a function, like the one given in our exercise, where you have a function, say \( f(x) = 2x^2 + 4x - 3 \).
The value of this function will change depending on the changes in \( x \). For example, if \( x \) changes from \( x_0 \) to \( x_0 + dx \), the change in the function's output is denoted as \( \Delta f \). This is calculated by finding the difference between \( f(x_0 + dx) \) and \( f(x_0) \).
  • For \( f(-0.9) \), we computed it to be \(-4.98\).
  • For \( f(-1) \), it was found to be \(-5\).
  • Thus, the change \( \Delta f \) is the difference between these values, yielding \( 0.02 \).
Understanding \( \Delta f \) helps us see how small increases or decreases in \( x \) influence the function's output.
Derivative
The derivative is a fundamental tool in calculus that allows us to understand how a function changes at any given point. It can be thought of as the function's rate of change: how one variable affects another.
For our function \( f(x) = 2x^2 + 4x - 3 \), the derivative, \( f'(x) \), is calculated as \( 4x + 4 \).
In this problem, we specifically evaluate the derivative at a point, \( x_0 = -1 \), giving us the value \( 0 \) (i.e., \( f'(-1) = -4 + 4 \)).
  • The derivative tells us that at \( x = -1 \), the function does not grow or decrease鈥攖he rate of change is zero, which informs us about the function's behavior at that point.
Calculating the derivative helps in approximating function values, especially for small changes, using differentials \( df \). Here, \( df \) is the product of the derivative and the small change \( dx \). For this exercise, \( df = f'(-1) \cdot 0.1 = 0 \cdot 0.1 = 0 \).
Approximation Error
When estimating the change in a function using derivatives, the approximation error is the difference between the actual change \( \Delta f \) and the estimated change using differentials \( df \). This helps in assessing how accurate our derivative-based approximation is.
In our example, \( \Delta f \) was calculated to be \( 0.02 \) and \( df \) was \( 0 \). The approximation error, therefore, is \( |0.02 - 0| = 0.02 \).
  • Approximation error quantifies the mismatch and shows how much the linear approximation diverges from the actual function value change.
  • Smaller errors indicate more accurate approximations.
Understanding approximation errors is key in judging the usefulness of derivatives in making predictions about function behavior.

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Most popular questions from this chapter

Graph \(y=-\sin x\) for \(-\pi \leq x \leq 2 \pi .\) On the same screen, graph $$ y=\frac{\cos (x+h)-\cos x}{h} $$ for \(h=1,0.5,0.3,\) and \(0.1 .\) Then, in a new window, try \(h=-1,-0.5,\) and \(-0.3 .\) What happens as \(h \rightarrow 0^{+} ?\) As \(h \rightarrow 0^{-9}\) What phenomenon is being illustrated here?

Find \(d y / d t\) when \(x=1\) if \(y=x^{2}+7 x-5\) and \(d x / d t=1 / 3\)

Find \(y^{\prime \prime}\) in Exercises \(59-64\) $$ y=\frac{1}{9} \cot (3 x-1) $$

Cardiac output In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Wurzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 \(\mathrm{L} / \mathrm{min.}\) . At rest it is likely to be a bit under 6 \(\mathrm{L} / \mathrm{min}\) . If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 \(\mathrm{L} / \mathrm{min.}\) Your cardiac output can be calculated with the formula $$y=\frac{Q}{D}$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{ml} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{ml} / \mathrm{min}\) and \(D=97-56=41 \mathrm{ml} / \mathrm{L}\) $$y=\frac{233 \mathrm{ml} / \mathrm{min}}{41 \mathrm{ml} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}$$ fairly close to the 6 \(\mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D. Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

A caution about centered difference quotients (Continuation of Exercise \(67 .\) ) The quotient $$ \frac{f(x+h)-f(x-h)}{2 h} $$ may have a limit as \(h \rightarrow 0\) when \(f\) has no derivative at \(x\) . As a case in point, take \(f(x)=|x|\) and calculate $$ \lim _{h \rightarrow 0} \frac{|0+h|-|0-h|}{2 h} $$ As you will see, the limit exists even though \(f(x)=|x|\) has no derivative at \(x=0 .\) Moral: Before using a centered difference quotient, be sure the derivative exists.
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