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Chapter 4: Problem 29

Use linear approximations to estimate the following quantities. Choose a value of a that produces a small error. \(1 / \sqrt[3]{510}\)

Short Answer

Expert verified
The approximate linear approximation for finding \(\frac{1}{\sqrt[3]{510}}\) is \(\frac{1}{8}+\frac{1}{3}\cdot 2^{-11}\).

Step by step solution

01

Find the function and its derivative

We need to find the derivative of the function \(f(x) = \frac{1}{\sqrt[3]{x}}\). First, rewrite the function as \(f(x) = x^{-1/3}\) to simplify the differentiation process. Now find the derivative: $$ f'(x) = -\frac{1}{3}x^{-4/3}. $$
02

Choose an appropriate value for a

We are asked to approximate \(\frac{1}{\sqrt[3]{510}}\). We must choose a value for \(a\) close to \(510\) for which we can easily find \(f(a)\). A close and easy-to-work-with value is \(a=512\), since it's a power of 2. Specifically, \(512 = 2^9\) and \(\sqrt[3]{512} = 2^3 =8\). So let's use \(a=512\).
03

Find \(f(a)\) and \(f'(a)\)

Now, we need to find \(f(512)\) and \(f'(512)\). We have: $$ f(512) = \frac{1}{\sqrt[3]{512}} = \frac{1}{8}, $$ and $$ f'(512) = -\frac{1}{3} (512)^{-4/3} = -\frac{1}{3}(2^9)^{-4/3} = -\frac{1}{3}(2^{-12}) = -\frac{1}{3}\cdot 2^{-12}. $$
04

Compute the linear approximation

We have all the required values to compute the linear approximation: $$ L(x) = f(a) + f'(a)(x-a) = \frac{1}{8} - \frac{1}{3}\cdot 2^{-12} (x - 512). $$ Now substitute \(x = 510\) to find our approximation: $$ L(510) = \frac{1}{8} - \frac{1}{3}\cdot 2^{-12} (510 - 512) = \frac{1}{8} + \frac{1}{3}\cdot 2^{-12} \cdot 2 = \frac{1}{8}+\frac{1}{3}\cdot 2^{-11}. $$ This gives us: $$\frac{1}{\sqrt[3]{510}} \approx \frac{1}{8}+\frac{1}{3}\cdot 2^{-11}.$$

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Key Concepts

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

Derivative Calculation
When we want to find the slope of a function at any given point, we must calculate its derivative. The exercise involves the function \(f(x) = \frac{1}{\sqrt[3]{x}}\), which can be rewritten as \(x^{-1/3}\) to make differentiation easier. This step transforms the problem into a more familiar form that follows basic power rule differentiation. The derivative of this function becomes:
  • \(f'(x) = -\frac{1}{3}x^{-4/3}\).
The derivative provides us with the rate of change of the function with respect to \(x\). This information is crucial in linear approximations because it helps us understand the behavior of the function near the point of interest. In this example, knowing \(f'(x)\) allows us to estimate how the value of \(\frac{1}{\sqrt[3]{510}}\) changes as \(x\) shifts slightly from an easy-to-calculate point.
Cubic Root Estimation
Estimating cubic roots involves finding values that are close, yet manageable enough to compute without complex calculations. In this practice problem, we needed to estimate \(\frac{1}{\sqrt[3]{510}}\). We chose a nearby number, 512, which makes calculations smoother because its cubic root is an integer. Specifically, 512 equals \(2^9\), and its cube root is \(2^3 = 8\).
The benefit of choosing nearby estimates, like 512 here, is that it's easier to work with integers and common powers, reducing inevitable calculation errors in real-world scenarios. Such estimations are extremely useful when precise instruments or computers are not accessible, enabling quick mental calculations that are sufficiently accurate for everyday purposes.
Function Approximation
Function approximation allows us to predict the behavior of complex functions around a specific point. Linear approximation is a common technique, using tangents of the function to make our predictions. By applying the formula:
  • \(L(x) = f(a) + f'(a)(x-a)\).
We substitute known values from our derivative calculation. Here, \(L(x)\) becomes the line that estimates \(f(x)\) near \(a\), where \(a\) is the value chosen near the point of interest.
In this case, linear approximation calculated \(L(510)\) by making use of the chosen \(a = 512\), considering both \(f(a) = \frac{1}{8}\) and \(f'(512)\), resulted in a very manageable approximation. These calculated approximations depict how a function behaves in small interval changes. This helps in quickly estimating complex calculations that would otherwise need higher computational resources.

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Most popular questions from this chapter

Given the following velocity functions of an object moving along a line, find the position function with the given initial position. Then graph both the velocity and position functions. $$v(t)=2 t+4 ; s(0)=0$$

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima. $$f(x)=2 x^{2} \ln x-11 x^{2}$$

Interpreting the derivative The graph of \(f^{\prime}\) on the interval [-3,2] is shown in the figure. a. On what interval(s) is \(f\) increasing? Decreasing? b. Find the critical points of \(f .\) Which critical points correspond to local maxima? Local minima? Neither? c. At what point(s) does \(f\) have an inflection point? d. On what interval(s) is \(f\) concave up? Concave down? e. Sketch the graph of \(f^{\prime \prime}\) f. Sketch one possible graph of \(f\)

Locate the critical points of the following functions and use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima. $$h(x)=(x+a)^{4} ; a \text { constant }$$

Basins of attraction Suppose \(f\) has a real root \(r\) and Newton's method is used to approximate \(r\) with an initial approximation \(x_{0} .\) The basin of attraction of \(r\) is the set of initial approximations that produce a sequence that converges to \(r .\) Points near \(r\) are often in the basin of attraction of \(r-\) but not always. Sometimes an initial approximation \(x_{0}\) may produce a sequence that doesn't converge, and sometimes an initial approximation \(x_{0}\) may produce a sequence that converges to a distant root. Let \(f(x)=(x+2)(x+1)(x-3),\) which has roots \(x=-2,-1\) and 3. Use Newton's method with initial approximations on the interval [-4,4] to determine (approximately) the basin of each root.
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