Problem 40 - 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 \(c\) $$ f(x)=x \ln (x), c=e $$

Step-by-Step Solution

Answer

The linearization at \( c = e \) is \( L(x) = 2x - e \).

Step 1: Find the Derivative of f(x)

First, we need to find the derivative of the function \( f(x) = x \ln(x) \). We use the product rule: if \( u = x \) and \( v = \ln(x) \), then the derivative is \( f'(x) = u'v + uv' = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1 \).

Step 2: Evaluate f(c) and f'(c)

Evaluate the function \( f(x) \) and its derivative \( f'(x) \) at \( c = e \). Thus, \( f(e) = e \ln(e) = e \cdot 1 = e \) and \( f'(e) = \ln(e) + 1 = 1 + 1 = 2 \).

Step 3: Setup the Linearization Formula

Now, we use the linearization formula \( L(x) = f(c) + f'(c) \cdot (x - c) \). Substitute \( f(e) = e \) and \( f'(e) = 2 \) into the equation: \( L(x) = e + 2 \cdot (x - e) \).

Step 4: Simplify the Linearization Equation

Simplify the linearization equation: \( L(x) = e + 2x - 2e = 2x - e \).

Key Concepts

Derivative

The concept of a derivative is a cornerstone of calculus. It represents the rate at which a function changes at any given point. Think of it as how fast something is moving at an exact instant. For the function \( f(x) = x \ln(x) \), the derivative \( f'(x) \) tells us how the output (\( f(x) \)) changes as its input (\( x \)) changes. In our example, finding the derivative involves understanding both the function \( f(x) \) and its components. We start by identifying parts of the function that can change: \( u = x \) and \( v = \ln(x) \). Calculating the derivatives of each part individually helps when applying rules like the product rule. Once we've calculated \( f'(x) \, = \ln(x) + 1 \, \), we know how the function behaves near any point.

Product Rule

When you're working with derivatives, you often deal with products of functions. The product rule is an essential technique in calculus that lets us differentiate such products.The product rule states that if you have two differentiable functions, say \( u \) and \( v \), then the derivative of their product \( uv \) is \( u'v + uv' \). This formula might look tricky at first, but it helps break down the process into manageable parts. For our function \( f(x) = x \ln(x) \), we identify \( u = x \) and \( v = \ln(x) \) such that the derivative is found using the product rule. The differentiation becomes \( 1 \cdot \ln(x) + x \cdot \frac{1}{x} \), which simplifies to \( \ln(x) + 1 \). This process helps us tackle more complex, combined functions in an organized way.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a mathematical function with unique properties that make it essential in calculus. It's the inverse of the exponential function, related to the constant \( e \) (approximately 2.718).When working with natural logarithms, remember their key qualities:

• The derivative of \( \ln(x) \) is \( \frac{1}{x} \).
• \( \ln(e) = 1 \) because \( e^1 = e \).

In our exercise, the natural logarithm allows us to understand changes in the function \( f(x) = x \ln(x) \). By differentiating \( \ln(x) \), we obtain \( \frac{1}{x} \), which simplifies calculations when using the product rule.

Calculus

Calculus is the branch of mathematics focusing on change and motion. It provides tools for understanding the dynamics of functions and models real-world phenomena. Central to calculus are the concepts of derivatives and integrals.In practical terms, calculus helps us:

• Understand instantaneous rates of change (derivatives).
• Calculate areas under curves (integrals), among other applications.

The original exercise involves using calculus concepts to linearize the function \( f(x) = x \ln(x) \) at \( c = e \). By using derivatives and evaluations at specific points, calculus offers insights into how a function behaves locally (near \( x = c \)). Here, the linearization \( L(x) \) gives a simple approximation of \( f(x) \) that is easier to calculate and understand.