Problem 24 - A Textbook of Differential Calculus

Question

Let \(f(x)\) be a polynomial with real coefficients such that \(f(x)=f^{\prime}(x) \times f^{\prime \prime}(x)\). If \(f(x)=0\) is satisfied \(x=1,2,3\) only, then the value of \(f^{\prime}(1) f^{\prime}(2) f^{\prime}(3)\) is (a) positive (b) negative (c) 0 (d) Inadequate data

Step-by-Step Solution

Answer

The value of \( f'(1)f'(2)f'(3) \) is negative.

Step 1: Understand Polynomial and Roots

The polynomial \( f(x) \) has roots \( x = 1, x = 2, \) and \( x = 3 \) only. Thus, \( f(x) \) can be expressed as \( f(x) = k(x-1)(x-2)(x-3) \), where \( k \) is a constant. The polynomial has real coefficients and is of degree 3 assuming no higher power in the polynomial.

Step 2: Find First and Second Derivatives

Calculate the first derivative, \( f'(x) \), and the second derivative, \( f''(x) \), based on \( f(x) = k(x-1)(x-2)(x-3) \). Use the product rule:- \( f'(x) = k((x-2)(x-3) + (x-1)(x-3) + (x-1)(x-2)) \)- Simplify \( f'(x) = k(3x^2 - 12x + 11) \). - \( f''(x) = k(6x - 12) \). These derivatives will be used to find \( f'(1), f'(2), \text{and } f'(3) \).

Step 3: Use Condition on Function and Derivatives

Given \( f(x) = f'(x) \times f''(x) \), we substitute the derivatives into the equation:\[ k(x-1)(x-2)(x-3) = k^2(3x^2 - 12x + 11)(6x - 12) \]This equation must hold for all \( x \), focusing especially on the roots.

Step 4: Evaluate Derivatives at Roots

Evaluate both the first and second derivatives at the roots:- \( f'(1) = 3k \), \( f''(1) = -6k \)- \( f'(2) = -3k \), \( f''(2) = 0 \)- \( f'(3) = 3k \), \( f''(3) = 6k \).

Step 5: Analyze Product of Derivatives at Roots

Evaluate the product \( f'(1) f'(2) f'(3) \):- \( f'(1) = 3k \), \( f'(2) = -3k \), \( f'(3) = 3k \).The product is \((3k) \times (-3k) \times (3k) = -27k^3 \). Therefore, as long as \( k eq 0 \), the product \( f'(1) f'(2) f'(3) \) is negative.

Key Concepts

Polynomial Roots

Roots of a polynomial are the values of \( x \) that make the polynomial equal to zero. In our exercise, the polynomial \( f(x) \) is defined to have real coefficients, meaning all the numbers are real, not complex.
Given roots are \( x = 1, 2, \) and \( 3 \), which implies that when \( x \) takes any of these values in \( f(x) \), the polynomial equals zero:

• \( f(1) = 0 \)
• \( f(2) = 0 \)
• \( f(3) = 0 \)

These points are where the graph of the polynomial intersects the x-axis. It helps us craft the polynomial expression:\( f(x) = k(x-1)(x-2)(x-3) \), with \( k \) being a constant determining the overall scale of the polynomial.

Derivative

The derivative of a function gives the rate at which the function's value changes as its input changes. It's like a tool to measure the slope or steepness of the function at any given point.
For our polynomial \( f(x) = k(x-1)(x-2)(x-3) \), the first derivative is calculated using the product rule. This rule involves differentiating each part and summing them up:
Using the product rule:

• Derivative: \( f'(x) = k((x-2)(x-3) + (x-1)(x-3) + (x-1)(x-2)) \)
• Simplified: \( f'(x) = k(3x^2 - 12x + 11) \)

At the roots, the derivative indicates how steep the polynomial is at those points, critical for comprehending how the polynomial behaves around its roots.

Second Derivative

The second derivative tells us about the curvature of the function—how the slope itself is changing. It helps identify the points where the function's graph changes from curving up to curving down, or vice versa.
So if we take the derivative of \( f'(x) = k(3x^2 - 12x + 11) \), we obtain the second derivative:

• Second Derivative: \( f''(x) = k(6x - 12) \)

In our exercise, we examine the second derivative values at the roots:

• At \( x = 1 \): \( f''(1) = -6k \)
• At \( x = 2 \): \( f''(2) = 0 \) (indicating a possible inflection point)
• At \( x = 3 \): \( f''(3) = 6k \)

This information aids in understanding the behavior of \( f(x) \) at these roots.

Real Coefficients

Real coefficients in a polynomial mean each term contains only real numbers. Such terms do not include imaginary or complex numbers, which are not used in this problem.
The polynomial equation in our exercise is \( f(x) = k(x-1)(x-2)(x-3) \). Here, \( k, 1, 2, \) and \( 3 \) are purely real numbers.

• Ensures the polynomial maintains its physical relevance in real-world applications.
• Stabilizes the behavior of the polynomial since complex roots do not exist.

Real coefficients help keep polynomials straightforward, while allowing us to focus on tangible roots and their derivatives, essential for evaluating expressions like \( f'(1) f'(2) f'(3) \) as seen in our solution.