/*! 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 12 Find a linearization at a suitab... [FREE SOLUTION] | 91Ó°ÊÓ

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Find a linearization at a suitably chosen integer near \(a\) at which the given function and its derivative are easy to evaluate. $$f(x)=\frac{x}{x+1}, \quad a=1.3$$

Short Answer

Expert verified
The linearization is \(L(x) = \frac{1}{4}x + \frac{1}{4}\) at \(x = 1\).

Step by step solution

01

Choose a Suitable Integer

To find the linearization of \(f(x) = \frac{x}{x+1}\) around \(a = 1.3\), first choose an integer near 1.3 where the function and its derivative are easy to compute. The closest integer to 1.3 is \(x = 1\).
02

Evaluate the Function at the Integer Point

Now, evaluate the function at \(x = 1\):\[ f(1) = \frac{1}{1+1} = \frac{1}{2} \]
03

Find the Derivative of the Function

Compute the derivative of the function \(f(x) = \frac{x}{x+1}\) using the quotient rule:\[ f'(x) = \frac{(1)(x+1) - x(1)}{(x+1)^2} = \frac{1}{(x+1)^2} \]
04

Evaluate the Derivative at the Integer Point

Evaluate the derivative at \(x = 1\):\[ f'(1) = \frac{1}{(1+1)^2} = \frac{1}{4} \]
05

Write the Linearization Formula

The linearization of the function around \(x = 1\) is given by:\[ L(x) = f(1) + f'(1)(x - 1) \]
06

Substitute Values Into the Linearization Formula

Substitute the values found for \(f(1)\) and \(f'(1)\) into the linearization formula:\[ L(x) = \frac{1}{2} + \frac{1}{4}(x - 1) \]Simplify the equation:\[ L(x) = \frac{1}{2} + \frac{1}{4}x - \frac{1}{4} = \frac{1}{4}x + \frac{1}{4} \]

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Key Concepts

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

Derivative
The derivative of a function provides information about its rate of change. When you calculate the derivative, you're essentially determining how a function behaves at any given point. In the context of the original exercise, we explored the function \( f(x) = \frac{x}{x+1} \). To obtain its derivative, the quotient rule is used since the function is a fraction of two expressions.

The derivative of \( f(x) \) using the quotient rule is calculated as follows:
  • The top part of the function is \( u = x \).
  • The bottom part of the function is \( v = x + 1 \).
  • The rule itself states: if \( f(x) = \frac{u}{v} \), the derivative \( f'(x) = \frac{u'v - uv'}{v^2} \).
  • For \( u = x \), the derivative \( u' = 1 \).
  • For \( v = x + 1 \), the derivative \( v' = 1 \).
  • Therefore, \( f'(x) = \frac{1 \cdot (x + 1) - x \cdot 1}{(x + 1)^2} = \frac{1}{(x+1)^2} \).
Understanding derivatives helps capture the way a function changes, which is essential for determining the linear approximation at a point.
Function Evaluation
In the process of finding a linearization, you need to evaluate the original function at a specific, convenient point. This step simplifies the linearization process and sets a base for approximating the function's behavior close to that point.

In our specific exercise, the function we deal with is \( f(x) = \frac{x}{x+1} \). To linearize, we evaluate this function at \( x = 1 \), which is the nearest integer to our given \( a = 1.3 \).
  • Plugging into the function:
    \( f(1) = \frac{1}{1+1} = \frac{1}{2} \).
This computed value is the part of the linear approximation formula \( L(x) = f(a) + f'(a)(x - a) \), once the derivative is also evaluated. The choice of a close integer simplifies computation, which is strategically significant in many linear approximation problems.
Quotient Rule
When you are dealing with a function that is a ratio of two other functions, the quotient rule is the method you use for finding the derivative. The rule is defined in such a way that it provides the derivative for functions in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions.

Following the quotient rule:
  • The derivative \( f'(x) = \frac{u'v - uv'}{v^2} \).
  • For \( f(x) = \frac{x}{x+1} \), \( u = x \) and \( v = x + 1 \).
  • Differentiate \( u \) to get \( u' = 1 \) and \( v \) to get \( v' = 1 \).
  • Substitute into the quotient rule to get \( f'(x) = \frac{1(x+1) - x \cdot 1}{(x+1)^2} \).
  • Simplifying gives \( f'(x) = \frac{1}{(x+1)^2} \).
The quotient rule is crucial for tackling complex rational expressions, like our function \( f(x) \), offering a coherent way to find their derivatives with ease.

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Most popular questions from this chapter

Which of the expressions are defined, and which are not? Give reasons for your answers. a. \(\csc ^{-1}(1 / 2)\) b. \(\csc ^{-1} 2\)

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find the equation for the tangent line to \(f\) at the specified point \(\left(x_{0}, f\left(x_{0}\right)\right)\) d. Find the equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right)\) Discuss the symmetries you see across the main diagonal. $$y=x^{3}-3 x^{2}-1, \quad 2 \leq x \leq 5, \quad x_{0}=\frac{27}{10}$$

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Use your graphing utility. Graph Newton's serpentine, \(y=4 x /\left(x^{2}+1\right)\) Then graph \(y=2 \sin \left(2 \tan ^{-1} x\right)\) in the same graphing window. What do you see? Explain.

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