Problem 12 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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 \(x = 1\).

Step 2: Evaluate the Function at the Integer Point

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

Step 3: Find the Derivative of the Function

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

Step 4: Evaluate the Derivative at the Integer Point

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

Step 5: Write the Linearization Formula

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

Step 6: 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:\[ L(x) = \frac{1}{2} + \frac{1}{4}x - \frac{1}{4} = \frac{1}{4}x + \frac{1}{4} \]

Key Concepts

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.