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. $$ f^{\prime \prime}(x)=2 \sqrt[3]{x+1} $$

Short answer

f(x)=\frac{9}{10}(x+1)^{7/3}+C_1x+C_2.

Step-by-step solution

Find the first derivative by antidifferentiating

To find the first derivative, we need to integrate the given expression for the second derivative, \( f''(x) = 2 \sqrt[3]{x+1} \). The antiderivative of \( \sqrt[3]{x+1} \) is \( \frac{3}{4} (x+1)^{\frac{4}{3}} \). Thus, after antidifferentiating, we obtain:\[ f'(x) = 2 \cdot \frac{3}{4} (x + 1)^{\frac{4}{3}} + C_1 = \frac{3}{2} (x + 1)^{\frac{4}{3}} + C_1 \]

Key concepts

Calculus

Calculus is a branch of mathematics that studies rates of change and the accumulation of quantities. It is divided into two main areas: differential calculus and integral calculus. Differential calculus focuses on finding the rate of change, which is often referred to as the derivative, while integral calculus deals with the accumulation of quantities, known as integration. Together, these concepts help us understand how functions change and how to calculate areas under curves, among many other applications.
When solving problems in calculus, particularly in differential equations, you might encounter higher-order derivatives, such as the second derivative you see in the problem. The goal in such exercises is often to reconstruct the original function by performing antidifferentiation, which is essentially integrating backwards. This process helps to find the function that produced the given rate of change, step by step, starting from the rates of change themselves.

Integration

Integration is a fundamental concept in calculus that involves finding a function from its derivative. In simpler terms, it's the process of reversing differentiation. To integrate a function, you find another function whose derivative is the original function. This process is crucial in calculus as it allows us to understand the total accumulation of a quantity, such as area under a curve, over an interval.
In the exercise example, the task was to find the original function from its second derivative, which requires integrating twice. The first integration provides the first derivative of the function, and the second integration uncovers the original function. Integration can involve various techniques such as substitution or integration by parts, depending on the form of the function to be integrated. Knowing basic derivatives helps considerably, as the integration of common functions often relies on reversing these basics. Always remember the integration will include an arbitrary constant since differentiation of a constant yields zero.

Arbitrary Constants

Arbitrary constants appear during the integration process due to the indefinite nature of antiderivatives. Since taking a derivative of a constant yields zero, any antiderivative must include an arbitrary constant, denoted as \(C\), to account for this fact.
In the context of the problem, when finding \(f(x)\) from \(f''(x)\), we encounter two arbitrary constants, \(C_1\) and \(C_2\), each arising from the two integrations performed. The first constant appears when finding \(f'(x)\), and the second emerges in the final expression for \(f(x)\). This reflects the fact that numerous functions share the same derivative but differ by a constant, making the role of arbitrary constants crucial to represent the family of possible original functions accurately.

• For every integration, expect one new arbitrary constant.
• These constants adjust the general solution to specific initial or boundary conditions when known.

Understanding these constants is essential for solving real-world problems, where constraints or initial values help determine their specific values.