Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(f^{\prime \prime}(x)\) is given. Find \(f(x)\) by antidifferentiating twice. Note that in this case your answer should involve two arbitrary constants, one from each antidifferentiation. For example, if \(f^{\prime \prime}(x)=x\), then \(f^{\prime}(x)=x^{2} / 2+C_{1}\) and \(f(x)=\) \(x^{3} / 6+C_{1} x+C_{2} .\) The constants \(C_{1}\) and \(C_{2}\) cannot be combined because \(C_{1} x\) is not a constant. $$ f^{\prime \prime}(x)=3 x+1 $$

Step-by-Step Solution

Answer

The function is \(f(x) = \frac{x^3}{2} + \frac{x^2}{2} + C_1 x + C_2\).

Step 1: Understand the Second Derivative

We are given the second derivative of a function, \(f''(x) = 3x + 1\). Our task is to find the original function \(f(x)\) by finding the antiderivatives.

Step 2: First Antiderivative

To find the first antiderivative, integrate \(f''(x) = 3x + 1\) with respect to \(x\). This gives:\[\int (3x + 1) \, dx = \frac{3x^2}{2} + x + C_1\]where \(C_1\) is the constant of integration. So, \(f'(x) = \frac{3x^2}{2} + x + C_1\).

Step 3: Second Antiderivative

Integrate \(f'(x) = \frac{3x^2}{2} + x + C_1\) to find \(f(x)\).\[\int \left(\frac{3x^2}{2} + x + C_1\right) \, dx = \frac{3x^3}{6} + \frac{x^2}{2} + C_1x + C_2\]Simplifying, we get:\[f(x) = \frac{x^3}{2} + \frac{x^2}{2} + C_1x + C_2\]where \(C_2\) is another constant of integration.

Key Concepts

Second Derivative

In calculus, the second derivative of a function gives us important information about the rate of change of its first derivative. When you see an expression like \(f''(x) = 3x + 1\), it indicates how the slope of the original function \(f(x)\) is changing. Understanding this helps you visualize how \(f(x)\) is shaped and how it behaves. The process of finding the original function \(f(x)\) from \(f''(x)\) involves the antidifferentiation process, performed twice to retrieve \(f(x)\).
In our example exercise, an antidifferentiation step applied to the second derivative will first yield the first derivative \(f'(x)\), and another will take us back to our original function \(f(x)\). Each integration step involves adding a new constant, reflecting indefinite integration. This relationship is vital in calculus for understanding the dynamics and behavior of functions.

Constant of Integration

When we integrate a function, we often encounter an unknown constant, known as the constant of integration. This constant is essential because integration is the reverse of differentiation, and during differentiation, any constant term becomes zero. Thus, to account for any possible constant term in the original function, we introduce this constant during the integration process.
The exercise demonstrates this by using two constants, \(C_1\) and \(C_2\), with each integration step. After finding the first antiderivative from \(f''(x)\), we introduce \(C_1\). Then, after finding the second antiderivative, another constant \(C_2\) is added. These constants represent the family of functions that can exist for the given second derivative, meaning there can be infinitely many solutions that differ by these constant terms.

Calculus Problems

Solving calculus problems often involves finding the original function from its derivatives. Such problems require knowledge of both differentiation and integration, and how these processes are interconnected.
When tackling a problem like the one provided, where the second derivative \(f''(x) = 3x + 1\) is given, you need to apply integration twice to determine \(f(x)\). Each step involves straightforward integration, but requires careful attention to include a constant of integration at each stage.

• Step 1: Integrate the second derivative to reveal the first derivative, introducing \(C_1\).
• Step 2: Integrate the first derivative to uncover the original function, introducing \(C_2\).

These steps will give you the complete expression for \(f(x)\), showcasing the importance of constants in integration.

Integration

Integration is a core operation in calculus, akin to the reverse process of differentiation. In the context of this problem, integration lets us work backward from a derivative to find the original function. Performing this operation twice on the given second derivative \(f''(x) = 3x + 1\) provides a clear method to recover \(f(x)\).
To carry out integration effectively, you'll:

• Identify the function components that require integration, such as polynomials and constant terms.
• Use the power rule and other integration techniques: for example, the power rule for integration states \(\int x^n \cdot dx = \frac{x^{n+1}}{n+1} + C\), which applies to the polynomial parts.

After each integration step, always remember to add a constant of integration. This practice ensures all possible forms of the original function are captured. The integration steps in the solution lead us to an expression for \(f(x)\) that incorporates both constants \(C_1\) and \(C_2\), the two constants arising from each step of integration.