Problem 41 - Calculus Single Variable

Question

Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=e(x-1) / x, c=1 $$

Step-by-Step Solution

Answer

The linearization of the function at \( c = 1 \) is \( L(x) = 1 \).

Step 1: Identify the Function and the Point of Interest

The function given is \( f(x) = \frac{e^{x-1}}{x} \), and we need to find the linearization at the point \( c = 1 \).

Step 2: Evaluate the Function at \( c \)

Substitute \( c = 1 \) into the function to find \( f(1) \):\[f(1) = \frac{e^{1-1}}{1} = \frac{e^0}{1} = 1.\]

Step 3: Find the Derivative of the Function

Use the quotient rule to differentiate \( f(x) = \frac{e^{x-1}}{x} \):The quotient rule is \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).Let \( u = e^{x-1} \) with \( u' = e^{x-1} \),and \( v = x \) with \( v' = 1 \).\[f'(x) = \frac{(e^{x-1})(x) - (e^{x-1})(1)}{x^2} = \frac{xe^{x-1} - e^{x-1}}{x^2} = \frac{e^{x-1}(x-1)}{x^2}.\]

Step 4: Evaluate the Derivative at \( c \)

Substitute \( c = 1 \) into \( f'(x) \):\[f'(1) = \frac{e^{1-1}(1-1)}{1^2} = \frac{e^0(0)}{1} = 0.\]

Step 5: Formulate the Linearization \( L(x) \)

The linearization formula is \( L(x) = f(c) + f'(c)(x - c) \). Substitute the values found:\[L(x) = 1 + 0(x - 1) = 1.\]

Key Concepts

Quotient Rule

The quotient rule is a fundamental tool in calculus used to find the derivative of a quotient of two functions. Suppose you have a function expressed as a ratio, say \( f(x) = \frac{u}{v} \), where \( u \) and \( v \) are functions of \( x \). To differentiate \( f \), we use the quotient rule:

• The formula for the quotient rule is: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).
• Here, \( u' \) is the derivative of \( u \), and \( v' \) is the derivative of \( v \).
• The numerator involves the difference between two products: \( u'v \) and \( uv' \).
• The denominator is simply \( v^2 \).

Using this rule helps to find derivatives when functions are divided, where direct differentiation is complex. In our example, with \( u = e^{x-1} \) and \( v = x \), the quotient rule simplifies the process of finding \( f'(x) \). Remember to carefully apply each step to avoid errors, especially noting the signs and placements of terms.

Differentiation

Differentiation is a process that estimates the rate at which a function changes. It's a crucial concept in calculus, particularly in determining slopes of tangent lines to curves. When you differentiate a function, you are essentially finding its derivative.

Purpose of Differentiation

• Calculates the rate of change or the slope at any point of the function.
• Essential in physics for motion and dynamics, and in economics for finding trends and optimizing functions.
• Helps in solving real-world problems where change is involved.

In the given exercise, differentiation involves using the quotient rule because of the division of two functions. Simplifying derivatives, like in our example where \( f(x) = \frac{e^{x-1}}{x} \), leads to understanding the behavior of functions at specific values.

Function Evaluation

Function evaluation involves substituting specific values into a function to determine its output. It's one of the fundamental methods for understanding how a function behaves at a certain point.

Steps in Function Evaluation

• Identify the point at which you need to evaluate the function, for instance, \( c = 1 \).
• Substitute this value into the function to find the result. For \( f(x) = \frac{e^{x-1}}{x} \), evaluating at \( x = 1 \) gives \( f(1) = 1 \).
• Ensure correct arithmetic: work systematically to avoid mistakes.

In the context of linearization, evaluating a function at a specific point \( c \) is the initial step for building the linear approximation \( L(x) \). This evaluation provides the function value needed to structure the linear equation, highlighting the significance of accuracy in these calculations. By determining \( f(c) \), in this case, 1, the groundwork for further calculations is properly set.