Problem 65 - Thomas Calculus

Question

Find the linearization of $$ f(x)=2-\int_{2}^{x+1} \frac{9}{1+t} d t $$ at \(x=1\)

Step-by-Step Solution

Answer

The linearization of \( f(x) \) at \( x = 1 \) is \( L(x) = -3x + 5 \).

Step 1: Understand the Function

The given function is \( f(x) = 2 - \int_{2}^{x+1} \frac{9}{1+t} \, dt \). It involves an integral from \( t = 2 \) to \( t = x + 1 \). To find the linearization, we'll need to take the derivative and evaluate it at \( x = 1 \).

Step 2: Differentiate the Function

To differentiate \( f(x) \), use the Fundamental Theorem of Calculus. If \( G(t) \) is the antiderivative of \( \frac{9}{1+t} \), then \( \frac{d}{dx} \left[ \int_{2}^{x+1} \frac{9}{1+t} \, dt \right] = \frac{9}{1+(x+1)} \cdot (1) \) due to the chain rule. Thus, \( f'(x) = -\frac{9}{x+2} \).

Step 3: Evaluate the Derivative at x=1

Now, substitute \( x = 1 \) into \( f'(x) = -\frac{9}{x+2} \). This results in \( f'(1) = -\frac{9}{3} = -3 \).

Step 4: Find the Function Value at x=1

Evaluate the function \( f(x) \) at \( x = 1 \): \( f(1) = 2 - \int_{2}^{2} \frac{9}{1+t} \, dt \). Since the integral's bounds are the same, its value is zero, implying \( f(1) = 2 \).

Step 5: Write the Linearization

The linearization of a function \( f \) at a point \( x = a \) is given by \( L(x) = f(a) + f'(a)(x-a) \). Substituting our values, we get \( L(x) = 2 - 3(x-1) \), which simplifies to \( L(x) = -3x + 5 \).

Key Concepts

Linearization

Linearization provides a simple way to approximate a complicated function using a straight line. Imagine zooming in closely on a curve at a particular point. Around this point, the curve can often be represented as a straight line. This line is the linearization.

In the context of the given exercise, the goal is to approximate the function \( f(x) = 2 - \int_{2}^{x+1} \frac{9}{1+t} \, dt \) near \( x = 1 \).

• The formula for linearization is \( L(x) = f(a) + f'(a)(x-a) \).
• This requires calculating both the function value \( f(a) \) and its derivative \( f'(a) \) at a specific point \( a \).
• For our exercise, we found that \( L(x) = -3x + 5 \), which approximates \( f(x) \) near \( x = 1 \).

This linear approximation helps in predicting the behavior of the function close to \( x = 1 \) without directly dealing with the complexities of the integral involved.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus serves as a bridge between differentiation and integration. It states that if you integrate a function and then differentiate the result, you will return to the original function.

This theorem has two main parts:

• The first part states that if \( F \) is the antiderivative of \( f \) on an interval \([a, b]\), then the integral of \( f \) from \( a \) to \( b \) is \( F(b) - F(a) \).
• The second part focuses on differentiation and tells us that if you differentiate the integral of a function with respect to its upper limit, you get the original function \( f(x) \).

In the exercise, to find the derivative \( f'(x) \) of the integral \( \int_{2}^{x+1} \frac{9}{1+t} \, dt \), we applied the second part of this theorem. We recognized that differentiating the integral with respect to \( x \) involved evaluating the integrand at the upper limit of the integral.

Integration

Integration is the process of finding the area under a curve. It is the reverse operation of differentiation. Various techniques exist for integration, including substitution, partial fractions, and integration by parts.

In this exercise, the integral \( \int_{2}^{x+1} \frac{9}{1+t} \, dt \) represents the area under the curve \( \frac{9}{1+t} \) from \( t = 2 \) to \( t = x+1 \). Here are some key points:

• Integration gives a cumulative total of quantities, such as areas, volumes, and accumulation over time.
• The integral in the problem has a variable upper bound, \( x+1 \), which indicates how the area changes as \( x \) changes.
• This setup is common in problems involving accumulated change, such as finding distance from velocity.

The techniques of integration allow us to handle functions like \( \frac{9}{1+t} \) that do not have simple antiderivatives.

Differentiation

Differentiation is the mathematical process of finding a derivative, which represents the rate of change of a function with respect to a variable. It is a fundamental concept in calculus.

To differentiate successfully, one must understand several rules and concepts:

• The Power Rule, Product Rule, and Chain Rule are essential tools for differentiating various types of functions.
• The Chain Rule is particularly crucial when dealing with composite functions.

In our problem, differentiation was applied using the Fundamental Theorem of Calculus and the Chain Rule.

We differentiated the integral expression, \( \int_{2}^{x+1} \frac{9}{1+t} \, dt \), using these concepts to find \( f'(x) = -\frac{9}{x+2} \). This derivative helped in determining the slope of the linear approximation at \( x = 1 \), pointing out how \( f(x) \) changes as \( x \) varies. Mastering differentiation enhances the understanding of how functions behave and change.