Problem 18 - 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 \(f(c)\) $$ f(x)=x e^{x}, c=0, x=0.12 $$

Step-by-Step Solution

Answer

Estimate: \( f(0.12) \approx 0.12 \).

Step 1: Identify f(c)

Since we are given the function \( f(x) = x e^x \) and that \( c = 0 \), we first find \( f(c) \). This is done by substituting \( c \) into the function.\[ f(0) = 0 \times e^0 = 0 \] So, \( f(c) = f(0) = 0 \).

Step 2: Calculate the Derivative f'(x)

To estimate \( f(x) \) at \( x = 0.12 \) using increments, we need the derivative \( f'(x) \). The derivative of \( f(x) = x e^x \) is found using the product rule.\[ f'(x) = (x)' e^x + x (e^x)' = e^x + x e^x = (1+x) e^x \]

Step 3: Evaluate f'(c)

Now, substitute \( c = 0 \) into \( f'(x) \) to find \( f'(c) \).\[ f'(0) = (1+0) e^0 = 1 \] So, \( f'(c) = f'(0) = 1 \).

Step 4: Use the Method of Increments

Using the method of increments (or linear approximations), we estimate \( f(x) \) as follows: \[ f(x) \approx f(c) + f'(c) \, (x - c) \] Substituting the known values: \[ f(0.12) \approx 0 + 1 \, (0.12 - 0) = 0.12 \]

Key Concepts

Increment Method

The increment method is a technique used to approximate the value of a function at a point close to where the value is known. It is beneficial in estimating values when calculating the exact outcome is difficult or cumbersome. Here's how it works in simple terms:

• First, identify a nearby value of the function that you know, such as \( f(c) \).
• Next, calculate the derivative of the function, \( f'(x) \), at that known value.
• Finally, apply the formula \( f(x) \approx f(c) + f'(c) \cdot (x - c) \) to estimate the function's value at an unknown point \( x \).

This approach is particularly effective when the changes in \( x \) are small, making the linear approximation fairly accurate.

Linear Approximation

Linear approximation is a method of estimating the value of a function using the tangent line at a point where the function value is known. This technique uses the first degree Taylor polynomial, leading to a simple linear formula.
The basic idea is to use the equation of the tangent line \( f(x) \approx f(c) + f'(c) \cdot (x-c) \). This requires two steps:

• Find the value of the function at a given point \( c \), which is \( f(c) \).
• Determine the slope of the function at that point, using the derivative \( f'(c) \).

By focusing on the tangent's linear relationship, this method allows you to make quick and straightforward approximations which are generally reasonable for slight changes in \( x \).

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of products of two functions. If you have a function \( f(x) = u(x) \cdot v(x) \), the product rule provides a formula to differentiate it:
\[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]
Here's a step-by-step breakdown of how the rule can be applied:

• Differentiate the first function, \( u(x) \), resulting in \( u'(x) \).
• Keep the second function, \( v(x) \), unchanged and multiply it by \( u'(x) \).
• Next, differentiate the second function, \( v(x) \), giving \( v'(x) \).
• Keep the first function, \( u(x) \), unchanged and multiply it by \( v'(x) \).
• Add these two resulting product terms together for the final derivative.

By enabling the differentiation of complex expressions like \( x e^x \), the product rule is essential for enhancing our ability to make accurate estimations and calculations.