Chapter 3: Problem 12
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 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, \( f'(x) = \frac{1 \cdot (x + 1) - x \cdot 1}{(x + 1)^2} = \frac{1}{(x+1)^2} \).
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 \( a = 1.3 \).
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:
\( f(1) = \frac{1}{1+1} = \frac{1}{2} \).
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:
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 \( f'(x) = \frac{1}{(x+1)^2} \).