Problem 247 - Calculus Volume 1

Question

For the following exercises, consider a third-degree polynomial \(f(x), \quad\) which has the properties \(f^{\prime}(1)=0, f^{\prime}(3)=0\) . Determine whether the following statements are true or false. Justify your answer. $$ f^{\prime \prime}(x)=0 \text { for some } 1 \leq x \leq 3 $$

Step-by-Step Solution

Answer

True: \(f''(x) = 0\) at \(x = 2\) within \(1 \leq x \leq 3\).

Step 1: Understand the Problem

We are given a third-degree polynomial \(f(x)\) with the derivative root at \(x = 1\) and \(x = 3\). We need to determine if there exists some \(x\) in the interval \(1 \leq x \leq 3\) where the second derivative \(f''(x) = 0\).

Step 2: Application of Rolle's Theorem

Rolle's Theorem states that if a function is continuous on a closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \(f(a) = f(b)\), then there is at least one \(c\) in \((a,b)\) such that \(f'(c) = 0\). Since \(f'(1) = 0\) and \(f'(3) = 0\), Rolle's Theorem suggests there must be a \(c\) in \(1 < c < 3\) where \(f''(c) = 0\).

Step 3: Analyzing the Derivative

Since \(f(x)\) is a third-degree polynomial, \(f'(x)\) is a second-degree polynomial. The solutions \(x = 1\) and \(x = 3\) suggest \(f'(x) = k(x - 1)(x - 3)\) for some constant \(k\) (i.e., \(f'(x)\) is a quadratic polynomial with roots at \(1\) and \(3\)).

Step 4: Deriving the Second Derivative

Differentiate \(f'(x) = k(x - 1)(x - 3)\) with respect to \(x\) to get the second derivative \(f''(x) = k(2x - 4)\).

Step 5: Solving for When \(f''(x) = 0\)

Set \(f''(x) = k(2x - 4) = 0\) and solve for \(x\):\[2x - 4 = 0 \x = 2\]

Step 6: Verify \(x = 2\) is in the Interval

Since \(x = 2\) lies within \(1 \leq x \leq 3\), the second derivative indeed equals zero for some \(x\) in the interval.

Key Concepts

Third-Degree Polynomial

A third-degree polynomial, often referred to as a cubic polynomial, is an algebraic expression of the form:\[ f(x) = ax^3 + bx^2 + cx + d \]where \( a, b, c, \) and \( d \) are constants and \( a eq 0 \). This polynomial is characterized by its cubic term, \( ax^3 \), which gives it a distinctive curve with up to three roots and two turning points, often forming an \'S\'-shaped curve.

Key Features:

• It has three roots, which are the x-values where the polynomial equals zero.
• It can have up to two turning points, which are peaks or valleys in the curve.
• The leading coefficient \( a \) affects the direction and width of the curve.

Understanding the properties of a third-degree polynomial is crucial when analyzing its derivatives, as these derivatives provide insight into the nature of the function's graph.

Second Derivative

The second derivative of a function like a third-degree polynomial provides information about the curvature or concavity of the graph. For a function \( f(x) \), the second derivative is denoted \( f''(x) \). It is the derivative of the first derivative, essentially describing how the slope of the curve changes.

Importance:

• A positive second derivative \( f''(x) > 0 \) indicates that the graph is concave up, or like a cup.
• A negative second derivative \( f''(x) < 0 \) indicates concave down, or like a cap.
• When \( f''(x) = 0 \), it could signal an inflection point, where the concavity changes.

In the context of the given problem, finding where \( f''(x) = 0 \) suggests that there might be an inflection point within the interval \([1, 3]\), as determined using Rolle's Theorem.

Quadratic Polynomial

A quadratic polynomial is a second-degree polynomial, which typically takes the form:\[ g(x) = ax^2 + bx + c \]where \( a, b, \) and \( c \) are constants and \( a eq 0 \). In the context of the exercise, when we derived the first derivative of the third-degree polynomial \( f(x) \), it resulted in a quadratic polynomial.

Characteristics:

• It has up to two roots and one vertex, which is either a minimum or maximum point.
• The coefficient \( a \) determines whether the parabola opens upward (\( a > 0 \)) or downward (\( a < 0 \)).
• The graph of a quadratic polynomial is a parabola.

In relation to the problem, since \( f'(x) \) is quadratic with roots at \( x = 1 \) and \( x = 3 \), it implies that the change in slope across these roots leads to identifying the x value where \( f''(x) = 0 \). This insight is vital for understanding where the concavity alters within the specified interval.