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Chapter 3: Problem 3

In Exercises \(1-4,\) find the linearization \(L(x)\) of \(f(x)\) at \(x=a\) $$ f(x)=x+\frac{1}{x}, \quad a=1 $$

Short Answer

Expert verified
The linearization of \(f(x) = x + \frac{1}{x}\) at \(x = 1\) is \(L(x) = 2\).

Step by step solution

01

Evaluate Function and Derivative

First, we evaluate the function \(f(x) = x + \frac{1}{x}\) at \(x = a = 1\), so \(f(1) = 1 + \frac{1}{1} = 2\). Next, find the derivative of the function: \(f'(x) = 1 - \frac{1}{x^2}\).
02

Evaluate the Derivative at a = 1

Evaluate the derivative at \(x = 1\): \(f'(1) = 1 - \frac{1}{1^2} = 1 - 1 = 0\).
03

Write the Linearization Formula

The linearization of a function \(f(x)\) at \(x = a\) is given by the formula \(L(x) = f(a) + f'(a)(x-a)\).
04

Find L(x)

Substitute the values we found into the formula: \(L(x) = 2 + 0 \cdot (x - 1) = 2\). So the linearization of \(f(x)\) at \(x = 1\) is \(L(x) = 2\).

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Key Concepts

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

Understanding Differential Calculus
Differential calculus is a fundamental branch of mathematics that focuses on how things change.
It鈥檚 grounded in the concept of derivatives, which measure how a function's output changes as its input changes.
At its core, differential calculus provides powerful tools for analyzing rates of change and slopes of curves.
By using derivatives, one can determine the instantaneous rate of change of a function, which can be highly useful in various real-world problems.
Whether it be physics, economics, or engineering, differential calculus allows us to make predictions and decisions based on how a small change in one quantity affects another.
  • Analyzes functions and their rates of change.
  • Uses derivatives to assess instantaneous change.
  • Widely applicable in science and engineering fields.
Introducing Derivatives
A derivative represents the rate at which a function is changing at any given point and is crucial in the study of calculus.
Imagine driving a car: the derivative tells you the car's speed at a particular moment, rather than the average speed over time.
In the context of the given problem, the function considered is \( f(x) = x + \frac{1}{x} \), and its derivative is \( f'(x) = 1 - \frac{1}{x^2} \).
The derivative measures how sensitive the function \( f(x) \) is to changes in \( x \), effectively providing a linear 鈥榮napshot鈥� of the slope of the curve.
  • Represents the slope of a function at a given point.
  • Provides a measure of how a function changes.
  • For \( f(x) = x + \frac{1}{x} \), \( f'(x) = 1 - \frac{1}{x^2} \).
Demystifying Linear Approximation
Linear approximation, also known as linearization, is a technique used to estimate the value of a function near a point using its tangent line.
It simplifies complex functions into linear forms, making them easier to work with and understand.
The formula for linearization at a point \( a \) is \( L(x) = f(a) + f'(a)(x - a) \).
In this exercise, the function \( f(x) = x + \frac{1}{x} \) is linearized at \( x = 1 \).
With \( f(1) = 2 \) and \( f'(1) = 0 \), the linear approximation formula becomes \( L(x) = 2 + 0 \cdot (x - 1) = 2 \).
Hence, at this specific point, the function is approximated to a constant value of 2.
  • Used to approximate function values near a specific point.
  • Transforms complicated functions into simpler linear ones.
  • In this problem, \( L(x) = 2 \), a constant approximation.

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Most popular questions from this chapter

a. Given that \(x^{4}+4 y^{2}=1\) , find \(d y / d x\) two ways: \((1)\) by solving for \(y\) and differentiating the resulting functions in the usual way and \((2)\) by implicit differentiation. Do you get the same result each way? b. Solve the equation \(x^{4}+4 y^{2}=1\) for \(y\) and graph the resulting functions together to produce a complete graph of the equation \(x^{4}+4 y^{2}=1 .\) Then add the graphs of the first derivatives of these functions to your display. Could you have predicted the general behavior of the derivative graphs from looking at the graph of \(x^{4}+4 y^{2}=1 ?\) Could you have predicted the general behavior of the graph of \(x^{4}+4 y^{2}=1\) by looking at the derivative graphs? Give reasons for your answers.

Is there anything special about the tangents to the curves \(y^{2}=x^{3}\) and \(2 x^{2}+3 y^{2}=5\) at the points \((1, \pm 1) ?\) Give reasons for your answer.

In Exercises \(67-70\) , use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\) . Perform the following steps: a. Plot the function \(f\) over \(I .\) b. Find the linearization \(L\) of the function at the point \(a\) . c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta > 0\) as you can, satisfying $$ |x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon $$ for \(\epsilon=0.5,0.1,\) and 0.01 . Then check graphically to see if your \(\delta\) -estimate holds true. $$ f(x)=\sqrt{x}-\sin x, \quad[0,2 \pi], \quad a=2 $$

Tangents Suppose that \(u=g(x)\) is differentiable at \(x=1\) and that \(y=f(u)\) is differentiable at \(u=g(1)\) . If the graph of \(y=f(g(x))\) has a horizontal tangent at \(x=1,\) can we conclude anything about the tangent to the graph of \(g\) at \(x=1\) or the tangent to the graph of \(f\) at \(u=g(1) ?\) Give reasons for your answer.

Use a CAS to perform the following steps on the parametrized curves in Exercises \(113-116 .\) a. Plot the curve for the given interval of \(t\) values. b. Find \(d y / d x\) and \(d^{2} y / d x^{2}\) at the point \(t_{0}\) . c. Find an equation for the tangent line to the curve at the point defined by the given value \(t_{0} .\) Plot the curve together with the tangent line on a single graph. $$ \begin{array}{l}{x=2 t^{3}-16 t^{2}+25 t+5, \quad y=t^{2}+t-3, \quad 0 \leq t \leq 6} \\ {t_{0}=3 / 2}\end{array} $$
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