Problem 16 - Calculus Single Variable

Question

Use the method of increments to estimate the value of \(f(x)\) at the given value of \(x\) using the known value \(f(c)\) $$ f(x)=x \ln (x), c=1, x=0.92 $$

Step-by-Step Solution

Answer

The estimated value of \( f(0.92) \) is \( -0.08 \).

Step 1: Understand the Function and Given Values

The function we are working with is \( f(x) = x \ln(x) \). We are given \( c = 1 \) where the function is known and want to estimate \( f(x) \) at \( x = 0.92 \). We know that \( f(1) = 1 \ln(1) = 0 \) since \( \ln(1) = 0 \).

Step 2: Set Up the Increment Approximation

The method of increments approximates functions near a known point \( c \). Using the linear approximation formula \( f(x) \approx f(c) + f'(c)(x-c) \), we need to find the derivative \( f'(x) \) first and then plug in the known values.

Step 3: Calculate the Derivative of f(x)

First, find \( f'(x) \). The derivative of \( f(x) = x \ln(x) \) can be found using the product rule: \( f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} \cdot 1 = \ln(x) + 1 \).

Step 4: Evaluate the Derivative at c

Substitute \( c = 1 \) into the derivative: \( f'(c) = \ln(1) + 1 = 0 + 1 = 1 \).

Step 5: Apply the Increment Formula

Using the formula from Step 2, evaluate \( f(x) \approx f(c) + f'(c)(x-c) \). We have \( f(1) = 0 \), \( f'(1) = 1 \), and \( x = 0.92 \). Thus, the approximation is: \[ f(0.92) \approx 0 + 1 \times (0.92 - 1) = 0.92 - 1 = -0.08 \].

Step 6: Final Estimate

The estimated value of \( f(0.92) \) is \( -0.08 \).

Key Concepts

Calculus

Calculus is an essential branch of mathematics that studies concepts of change and motion. Unlike static mathematics disciplines, calculus gives us tools to understand how quantities vary. By examining rates of change and accumulations of quantities, we can analyze dynamic systems.
Through concepts like limits, derivatives, and integrals, calculus helps us understand complex relationships in mathematical contexts.

• Derivatives: Study the rate of change of a function.
• Integrals: Analyze accumulated changes over an interval.
• Limits: Explore behavior as values approach a certain point.

When using mathematical approximations such as the linear approximation above, calculus provides a solid framework to make accurate estimations in the vicinity of known points.

Derivative

Derivatives are a core concept in calculus referring to the rate at which a function changes at any given point. To find the derivative of a function, we use rules such as the product rule and chain rule.
In the current exercise, the function is given as \( f(x) = x \ln(x) \). Using the product rule, which applies when differentiating a product of two functions, we find:\[ f'(x) = \ln(x) + 1 \] The derivative explains how \( f(x) \) rises or falls as \( x \) changes, offering insight into the function's behavior near \( x = 1 \).

• Product Rule: If you have a function \( f(x) = u(x) \cdot v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \).
• The derivative provides a linear estimation that can be used for approximations, such as in the increment method.

Function Approximation

Function approximation is the process of finding a simple function that is close to the given function, especially useful when the exact function is complex or unknown at certain points. Linear approximation is a common approximation method that uses tangent lines to estimate values nearby a known point. In simpler terms, near a point \(c\), a function can be approximated as a straight line.
The formula for linear approximation is:

• \( f(x) \approx f(c) + f'(c)(x-c) \)
• It uses the value of the function and its derivative at a known point \(c\) to estimate \(f(x)\).

In the exercise, the linear approximation helped estimate \( f(0.92) \) by using a known value at \( c = 1 \). By simplifying calculations, it provides a quick way to understand how the function behaves in a small interval around \( c \).

Increment Method

The increment method is a technique in calculus used for estimating the change in a function's value, drawing from the linear approximation principle. This method is particularly useful when working with small deviations from a known point. In essence, it assumes the change is linear and directly proportional to the increment in \(x\).

• The general formula is \( f(x) \approx f(c) + f'(c) \, (x-c) \).
• For small \(x\) changes, like \(x = 0.92\) from \(x = 1\), this method offers quick approximate evaluations.

By using this method, you find that even without computing \(f(x)\) exactly, you can still gain valuable insights and approximations for the function's behavior, making it a powerful tool in both theory and practical applications.