Problem 26 - Calculus with Concepts in Calculus

Question

Determine all functions \(f\) satisfying the given conditions. $$ f^{(3)}(x)=0 \text { (Hint : Use Theorem 4.6 three times.) } $$

Step-by-Step Solution

Answer

The functions satisfying \( f^{(3)}(x) = 0 \) are quadratic polynomials: \( f(x) = \frac{C_1}{2}x^2 + C_2x + C_3 \).

Step 1: Understanding the Problem

The problem asks us to find all functions \( f(x) \) whose third derivative is zero, i.e., \( f^{(3)}(x) = 0 \). According to calculus, if the third derivative of a function is zero, the function is a polynomial of degree at most 2, as higher derivatives of polynomials lower the degree by one each time.

Step 2: Applying Theorem 4.6

Theorem 4.6 likely refers to the Fundamental Theorem of Calculus or another theorem relating to derivatives. Applying it three times indicates integrating the third derivative, \( f^{(3)}(x) = 0 \), back down to the original function.

Step 3: First Integration

Integrate \( f^{(3)}(x) = 0 \) to get \( f''(x) = C_1 \), where \( C_1 \) is a constant of integration. This is because the derivative of a constant is zero, hence integrating restores the constant.

Step 4: Second Integration

Integrate \( f''(x) = C_1 \) once more to obtain \( f'(x) = C_1x + C_2 \), where \( C_2 \) is another constant of integration. This step represents finding the original function's first derivative.

Step 5: Third Integration

Finally, integrate \( f'(x) = C_1x + C_2 \) to find \( f(x) = \frac{C_1}{2}x^2 + C_2x + C_3 \), with \( C_3 \) being another constant of integration. This yields a quadratic polynomial for \( f(x) \).

Step 6: Verifying the Solution

Verify that the function \( f(x) = \frac{C_1}{2}x^2 + C_2x + C_3 \) has a third derivative of zero. Calculating \( f^{(3)}(x) \) confirms it is zero, as differentiation reduces constants to zero.

Key Concepts

Polynomial Functions

Polynomial functions are expressions constructed from variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents. These functions are fundamental in algebra and calculus due to their straightforward properties and predictable behaviors.
Polynomials are usually written in the general form as: \[p(x) = a_nx^n + a_{n-1}x^{n-1} + \ a_{n-2}x^{n-2} + ... + a_1x + a_0\]where:

• \( n \) is a non-negative integer representing the degree of the polynomial.
• \( a_n, a_{n-1}, ..., a_1, a_0 \) are constants, known as coefficients.

The degree of a polynomial determines its general shape and complexity. In the context of the exercise, since the third derivative \( f^{(3)}(x) \) is zero, it implies that the polynomial function could at most have a degree of 2. Higher degree terms fall away due to repeated differentiation.

Integration

Integration is the reverse process of differentiation. It involves finding a function whose derivative is the given function. This process is essential in calculating areas, volumes, and other quantities defined over continuous intervals.
For polynomials, integration is straightforward. When you integrate a function repeatedly, constants are introduced at every step, called constants of integration. Given a function's derivative, you can find the primitive function (original function) through integration.

• The integral of a constant \( c \) is \( cx + C \), where \( C \) is the constant of integration.
• For a function like \( ax^n \), its integral becomes \( \frac{a}{n+1}x^{n+1} + C \) for \( n eq -1 \).

In our exercise, starting from \( f^{(3)}(x) = 0 \), successive integrations returned the functional form of \( f(x) = \frac{C_1}{2}x^2 + C_2x + C_3 \), representing the cumulative constants of integration incorporated.

Calculus Theorems

Calculus theorems provide essential rules for differentiation and integration, allowing us to solve complex problems by breaking them into manageable steps. Although the original exercise references a specific Theorem 4.6, its essence revolves around fundamental concepts like:

• The Fundamental Theorem of Calculus, which links the concept of differentiation and integration, stating that integration can reverse differentiation and vice versa.
• Rules for differentiating and integrating different types of functions, providing systematic approaches for both operations.

In our particular problem, the third derivative being zero tells us that integrating multiple times will yield a polynomial function characterized by its constant terms. This idea, supported indirectly by the calculus theorems, confirms why multiple steps of integration were used to determine the original function.