Chapter 5: Problem 65
Find the linearization of $$ f(x)=2-\int_{2}^{x+1} \frac{9}{1+t} d t $$ at \(x=1\)
Short Answer
Expert verified
The linearization of \( f(x) \) at \( x = 1 \) is \( L(x) = -3x + 5 \).
Step by step solution
01
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 \( x = 1 \).
02
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, \( f'(x) = -\frac{9}{x+2} \).
03
Evaluate the Derivative at x=1
Now, substitute \( x = 1 \) into \( f'(x) = -\frac{9}{x+2} \). This results in \( f'(1) = -\frac{9}{3} = -3 \).
04
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 \( f(1) = 2 \).
05
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 \( L(x) = -3x + 5 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( x = 1 \).
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 \( x = 1 \).
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:
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 \( f(x) \).
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:
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.
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:
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.
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.
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.