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Use linear approximations to estimate the following quantities. Choose a value of a that produces a small error. \(1 / 203\)

Short Answer

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Question: Use linear approximation to estimate the value of \(\frac{1}{203}\). Answer: The linear approximation estimate for \(\frac{1}{203}\) is \(\frac{39400}{8000000}\).

Step by step solution

01

Choose a function and a value for a

Let's consider the function \(f(x) = \frac{1}{x}\). We want to find a value of \(a\) near \(203\) so that \(f(a) \approx \frac{1}{203}\). We'll choose \(a = 200\) because it is close to \(203\) and has a simpler representation than \(203\) itself.
02

Calculate the derivative of the function

We will now find the derivative of the function \(f(x) = \frac{1}{x}\) with respect to \(x\). Using the power rule for differentiation, we have: \[f'(x) = -\frac{1}{x^2}\]
03

Evaluate the derivative at a

Now, we need to evaluate the derivative at \(a = 200\). We have: \[f'(200) = -\frac{1}{200^2} = -\frac{1}{40000}\]
04

Use the linear approximation formula

The linear approximation formula is given by: \[f(x) \approx f(a) + f'(a) (x - a)\] Plugging in the values we've obtained so far: \[\frac{1}{203} \approx f(203) \approx f(200) + f'(200)(3)\] \[f(203) \approx \frac{1}{200} - \frac{1}{40000}(3)\]
05

Calculate the estimate

Finally, we will calculate the estimate using the linear approximation formula: \[\frac{1}{203} \approx \frac{1}{200} - \frac{1}{40000}(3)\] \[\frac{1}{203} \approx \frac{1}{200} - \frac{3}{40000}\] \[\frac{1}{203} \approx \frac{40000 - 3\cdot 200}{40000\cdot200}\] \[\frac{1}{203} \approx \frac{40000 - 600}{8000000}\] \[\frac{1}{203} \approx \frac{39400}{8000000}\] This is our linear approximation estimate for \(\frac{1}{203}\).

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Key Concepts

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

Derivative Calculation
Understanding how to calculate derivatives is a foundational skill in calculus. The derivative represents the rate at which a function's output changes with respect to a change in its input. Essentially, it's like the slope of the tangent line at any point on a graph of the function.

When we computed the derivative of the function \(f(x) = \frac{1}{x}\), what we were really finding out is how the function value changes as we make tiny movements along the x-axis. In this case, the derivative \(f'(x)\) tells us that for every unit we move away from a specific point \(a\), the function is changing at a rate proportional to \(\frac{1}{x^2}\).

This information is crucial when we want to estimate function values without calculating them exactly, which is the essence of the linear approximation method we employed to estimate \(\frac{1}{203}\).
Power Rule for Differentiation
The power rule for differentiation is an easy yet powerful tool to find the derivative of functions of the form \(x^n\), where \(n\) is any real number. The rule simply states that the derivative of \(x^n\) is \(nx^{n-1}\).

For our exercise, \(f(x) = \frac{1}{x} = x^{-1}\), so applying the power rule gives us \(f'(x) = -1\cdot x^{-1-1} = -\frac{1}{x^2}\). This straightforward technique bypasses the need for complex derivative calculations, thus simplifying our work significantly.

The accuracy of derivatives obtained using the power rule is indispensable for making reliable linear approximations, as it directly affects the slope of the line that approximates the function near a given point.
Linear Approximation Formula
The linear approximation formula is a way to estimate the value of a function near a point \(a\) using its derivative. It states that \(f(x) \approx f(a) + f'(a) \cdot (x - a)\), where \(f'(a)\) is the derivative of \(f\) at \(a\). This is essentially the equation of the tangent line at \(a\), which we use as an estimate of \(f\) near \(a\).

In the given exercise, by using \(a = 200\), which is close to \(203\), we utilized this formula to find an approximated value of \(\frac{1}{203}\) based on the known value of \(\frac{1}{200}\). By adding the change (the product of the slope \(f'(a)\) and the distance \(x - a\)), we effectively 'walk' along the tangent line from \(a\) to our point of interest, \(203\), achieving an estimate close to the real value.

This powerful method is particularly useful when dealing with functions that are complex to evaluate directly but can be approximated near points where the function's value is known or easy to calculate.

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