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Find \(f\) by solving the initial value problem. $$ f^{\prime \prime}(x)=\frac{1}{x^{3}} ; \quad f(1)=1, \quad f^{\prime}(1)=\frac{1}{2} $$

Short Answer

Expert verified
The function \(f(x)\) is found by solving the initial value problem, resulting in \(f(x) = \frac{1}{x} + x - 1\).

Step by step solution

01

Integrating the second derivative to find the first derivative

First, we must integrate the given second derivative, \(f^{\prime \prime}(x) = \frac{1}{x^3}\), with respect to \(x\) to obtain the first derivative \(f^{\prime}(x)\). $$ f^{\prime}(x) = \int f^{\prime \prime}(x) dx = \int \frac{1}{x^3} dx $$
02

Solve the integral

To solve this integral, we can rewrite the integrand as \(x^{-3}\) and perform the integration: $$ f^{\prime}(x) = \int x^{-3} dx = \frac{x^{-3+1}}{-3+1} + C_1 = \frac{x^{-2}}{-2} + C_1 = -\frac{1}{2x^2} + C_1 $$
03

Apply the initial condition to find \(C_1\)

We are given the initial condition \(f^{\prime}(1) = \frac{1}{2}\). We'll substitute this into our expression for \(f^{\prime}(x)\) to find the value for \(C_1\). $$ \frac{1}{2} = -\frac{1}{2(1)^2} + C_1 $$ Solve for \(C_1\): $$ C_1 = \frac{1}{2} + \frac{1}{2} = 1 $$ Our expression for \(f^{\prime}(x)\) is now: $$ f^{\prime}(x) = -\frac{1}{2x^2} + 1 $$
04

Integrating the first derivative to find the function \(f(x)\)

Now, we need to integrate the first derivative, \(f^{\prime}(x) = -\frac{1}{2x^2} + 1\), with respect to \(x\) to obtain the original function \(f(x)\). $$ f(x) = \int f^{\prime}(x) dx = \int \left(-\frac{1}{2x^2} + 1\right) dx $$
05

Solve the integral

Split the integral and solve each term separately: $$ f(x) = \int \left(-\frac{1}{2x^2}\right) dx + \int 1 \space dx = -\frac{1}{2}\int x^{-2} dx + \int 1 \space dx $$ Integrate each term: $$ f(x) = -\frac{1}{2} \frac{x^{-2 + 1}}{-2+1} + x + C_2 = \frac{1}{x} + x + C_2 $$
06

Apply the initial condition to find \(C_2\)

We are given the initial condition \(f(1) = 1\). We'll substitute this into our expression for \(f(x)\) to find the value for \(C_2\). $$ 1 = \frac{1}{1} + 1 + C_2 $$ Solve for \(C_2\): $$ C_2 = 1 - 2 = -1 $$
07

Write the final solution

The solution to the problem is the function \(f(x)\), which can now be written as $$ f(x) = \frac{1}{x} + x - 1 $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Integration is a fundamental tool in calculus that allows us to find the original function, known as the antiderivative, given its derivative. In the context of initial value problems, integration is used to reconstruct the original function from its derivatives.

Consider the exercise provided; we are given the second derivative of a function and are tasked with finding the original function. This process involves integrating the second derivative to obtain the first derivative, and then integrating the first derivative to get the original function. These integrals are evaluated step by step, applying reverse operations of differentiation.

Why is Integration Important?
  • It helps in solving differential equations.
  • Integration provides the area under curves, which is fundamental in physics and engineering.
  • It is crucial for finding displacement given velocity or acceleration.
Understanding how to solve integrals, especially when they involve a simple power of x, is key. The integral of a function such as \(x^{-n}\) is found by using the formula \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), where \(C\) is the integration constant. This formula is applied in the steps provided to reach the solution for the first and the original function, \(f(x)\).
Second Derivative
The second derivative of a function is the derivative of the derivative. It provides information about the concavity of the function and how the rate of change of the function's slope is varying.

In the exercise at hand, the second derivative, expressed as \(f''(x)\), gives us the rate at which the first derivative \(f'(x)\) is changing. When you're solving an initial value problem, starting with the second derivative often indicates that you will be working your way back to the original function by integrating twice.

The Significance of the Second Derivative:
  • It tells us if a function is concave up or down at a particular point.
  • It is essential for finding points of inflection where the concavity changes.
  • It is used in optimization problems to determine the nature of critical points (maxima or minima).
By integrating the given second derivative, we step closer to finding the original function, which is the aim of the exercise.
Initial Conditions
Initial conditions are specific values given for a function or its derivatives at a particular point, and they allow us to find the unique solution of a differential equation. Without initial conditions, the solution could be any one of an infinite number of possible functions.

In our example, the initial conditions \(f(1) = 1\) and \(f'(1) = \frac{1}{2}\) provide essential information that we use to determine the constants of integration, \(C_1\) and \(C_2\), in the integration process. After integrating to find the general form of the first derivative and the original function, we apply these conditions to find the specific solution that satisfies the given problem.

The Role of Initial Conditions:
  • They make our solution to a differential equation unique.
  • Initial conditions are crucial in applications such as physics and engineering, where they represent physical quantities at a starting time.
  • Without initial conditions, we cannot fully apply integration to find a specific solution to practical problems.
Applying these given conditions at the right step is crucial; for the first derivative, we use \(f'(1)\) after the first integration, and for the original function, \(f(1)\) after integrating the first derivative.

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