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Magazine Chapter 4: Problem 45
\(23-46\) Find \(f\) $$f^{\prime \prime}(x)=x^{-2}, \quad x>0, \quad f(1)=0, \quad f(2)=0$$
Short Answer
Expert verified
The function is \(f(x) = -\ln|x| + \ln(2)x - \ln(2)\).
Step by step solution
01
Integrate the Second Derivative
We begin by finding the first derivative from the second derivative. Given that \(f''(x) = x^{-2}\), we integrate with respect to \(x\) to find \(f'(x)\). The integral of \(x^{-2}\) is:\[\int f''(x) \, dx = \int x^{-2} \, dx = -x^{-1} + C_1\]Therefore, \(f'(x) = -x^{-1} + C_1\).
02
Integrate the First Derivative
Now, find \(f(x)\) by integrating \(f'(x)\). The first derivative is \(f'(x) = -x^{-1} + C_1\), so we integrate:\[\int f'(x) \, dx = \int (-x^{-1} + C_1) \, dx = -\ln|x| + C_1x + C_2\]Thus, \(f(x) = -\ln|x| + C_1x + C_2\).
03
Apply Initial Conditions
We use the initial conditions \(f(1) = 0\) and \(f(2) = 0\) to solve for \(C_1\) and \(C_2\).1. For \(f(1) = 0\): \[-\ln(1) + C_1(1) + C_2 = 0\] Since \(\ln(1) = 0\), this simplifies to: \[C_1 + C_2 = 0\] 2. For \(f(2) = 0\): \[-\ln(2) + C_1(2) + C_2 = 0\] \[C_1(2) + C_2 = \ln(2)\]Solving these two equations, we find: \[C_1 = \ln(2), \; C_2 = -\ln(2)\].
04
Write the Final Function
Substituting back \(C_1\) and \(C_2\) into the expression for \(f(x)\), we have:\[f(x) = -\ln|x| + \ln(2)x - \ln(2)\] The function \(f(x)\) satisfies the given conditions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second Derivative
In calculus, the second derivative of a function provides significant insights into the behavior of the original function. It is essentially the derivative of the first derivative, which means it tells us how the rate of change itself is changing. For a given function, if the second derivative is positive, the original function is concave up, indicating that it has a local minimum, while a negative second derivative indicates a concave down shape, showing a local maximum.
To solve problems involving the second derivative, like the one in the exercise, we first start by reverse engineering the process鈥攊ntegrating it to find the first derivative. This not only helps in recovering the original function but also in understanding how the function behaves with respect to continuous changes in the inputs, especially around the critical points.
Learning to manipulate and think about second derivatives is a crucial step in deepening your understanding of calculus and applying it to real-world problems.
To solve problems involving the second derivative, like the one in the exercise, we first start by reverse engineering the process鈥攊ntegrating it to find the first derivative. This not only helps in recovering the original function but also in understanding how the function behaves with respect to continuous changes in the inputs, especially around the critical points.
Learning to manipulate and think about second derivatives is a crucial step in deepening your understanding of calculus and applying it to real-world problems.
Initial Conditions
Initial conditions are like clues that help us unlock the specifics of a function. When a function is derived through integration, it usually contains unknown constants. Initial conditions are provided to eliminate these unknowns and find the specific solution that fits the problem. In our exercise, the initial conditions given are
By substituting the values from initial conditions into the function, we create a system of equations that can be solved to find these constants. In the exercise, using these conditions sequentially brought us to specific values for constants \(C_1\) and \(C_2\), ensuring that the reconstructed function satisfies all given requirements.
- \(f(1) = 0\)
- \(f(2) = 0\)
By substituting the values from initial conditions into the function, we create a system of equations that can be solved to find these constants. In the exercise, using these conditions sequentially brought us to specific values for constants \(C_1\) and \(C_2\), ensuring that the reconstructed function satisfies all given requirements.
Antiderivative
Finding an antiderivative is essentially the process of integration. It is the reverse of differentiation and provides a function whose derivative matches the original given function. The challenge often lies not just in finding an antiderivative, but in finding the correct one that fits the context of a problem with initial or boundary conditions.
For example, in our specific exercise, we were given a second derivative and we needed to integrate it twice to uncover the original function. First, integrating the second derivative provided us with the first derivative. From there, a further integration revealed the original function \(f(x)\) with additional constants, which needed to be determined using given conditions.
Understanding how to integrate effectively and the role of antiderivatives is foundational in both solving differential equations and in conceptualizing real-world phenomena as they evolve naturally over time.
For example, in our specific exercise, we were given a second derivative and we needed to integrate it twice to uncover the original function. First, integrating the second derivative provided us with the first derivative. From there, a further integration revealed the original function \(f(x)\) with additional constants, which needed to be determined using given conditions.
Understanding how to integrate effectively and the role of antiderivatives is foundational in both solving differential equations and in conceptualizing real-world phenomena as they evolve naturally over time.
Constant of Integration
In any indefinite integral, a constant of integration is added to account for all possible vertical shifts of the antiderivative function. This constant is vital because differentiation obliterates constants, meaning the original function could have included any constant value.
When you encounter a problem with initial conditions, these constants are not arbitrary. Instead, they must be evaluated so the solution fits the conditions specified in the problem. To determine these constants, like in our exercise, use the initial conditions given to set up equations that will isolate each constant. This method ensures your final answer isn't just any antiderivative, but the specific one that meets the constraints like \(f(1) = 0\) and \(f(2) = 0\).
Recognizing the significance of the constant of integration goes a long way in solving not only textbook problems but also in verifying answers when modeling scenarios that depend on specific starting conditions.
When you encounter a problem with initial conditions, these constants are not arbitrary. Instead, they must be evaluated so the solution fits the conditions specified in the problem. To determine these constants, like in our exercise, use the initial conditions given to set up equations that will isolate each constant. This method ensures your final answer isn't just any antiderivative, but the specific one that meets the constraints like \(f(1) = 0\) and \(f(2) = 0\).
Recognizing the significance of the constant of integration goes a long way in solving not only textbook problems but also in verifying answers when modeling scenarios that depend on specific starting conditions.
