Problem 38 - Applied Calculus

Question

A cubic polynomial, \(a x^{3}+b x^{2}+c x+d,\) with a critical point at \(x=2,\) an inflection point at \((1,4),\) and a leading coefficient of 1.

Step-by-Step Solution

Answer

The polynomial is \( f(x) = x^3 - 3x^2 + 6 \).

Step 1: Understanding the Problem

We need to find a cubic polynomial of the form \( f(x) = ax^3 + bx^2 + cx + d \) with a leading coefficient of 1. The critical point given is \( x = 2 \) and the inflection point is at \((1, 4)\).

Step 2: Setting the Leading Coefficient

We are given that the leading coefficient of the polynomial is 1, so \( a = 1 \). Thus, the polynomial simplifies to \( f(x) = x^3 + bx^2 + cx + d \).

Step 3: Finding the Derivative

Calculate the first derivative of \( f(x) = x^3 + bx^2 + cx + d \).\[ f'(x) = 3x^2 + 2bx + c \]

Step 4: Using the Critical Point Condition

The critical point at \( x = 2 \) means that \( f'(2) = 0 \). Substitute \( x = 2 \) into the derivative:\[ 3(2)^2 + 2b(2) + c = 0 \]Simplifying, we get \[ 12 + 4b + c = 0 \] or \( 4b + c = -12 \). This is Equation (1).

Step 5: Finding the Second Derivative

Calculate the second derivative of \( f(x) \):\[ f''(x) = 6x + 2b \]

Step 6: Using the Inflection Point Condition

An inflection point at \( x = 1 \) requires that \( f''(1) = 0 \). Substitute \( x = 1 \) into the second derivative:\[ 6(1) + 2b = 0 \]Simplifying, we find \( 6 + 2b = 0 \), which means \( b = -3 \).

Step 7: Solving for 'c' using Equation (1)

Now substitute \( b = -3 \) into Equation (1):\[ 4(-3) + c = -12 \]\[ -12 + c = -12 \]Thus, \( c = 0 \).

Step 8: Finding 'd' using the Polynomial Value at the Inflection Point

Substitute \( x = 1 \) and \( y = 4 \) into the polynomial:\[ f(1) = 1^3 - 3(1)^2 + 0(1) + d = 4 \]\[ 1 - 3 + d = 4 \]\[ d = 6 \].

Step 9: Finalizing the Polynomial

The coefficients found are \( a = 1 \), \( b = -3 \), \( c = 0 \), and \( d = 6 \). Therefore, the final polynomial is:\[ f(x) = x^3 - 3x^2 + 6 \].

Key Concepts

Understanding Critical Points

Critical points of a polynomial are where the derivative is zero, indicating potential peaks, troughs, or flat areas. For a cubic polynomial, like the one in our problem, there could be at most two real critical points. In our scenario, the critical point is at \( x = 2 \). This means that if we were to graph the polynomial, at \( x = 2 \), the slope of the tangent will be zero, representing either a local maximum or minimum. To find a critical point, set the first derivative of the polynomial equal to zero and solve for \( x \). Understanding this helps in determining the shape and behavior of the polynomial at specific points on its graph.

Exploring Inflection Points

Inflection points occur where the second derivative of a polynomial changes sign. This indicates a change in the curve's concavity. For cubic polynomials, typically, there's one inflection point. In this problem, the inflection point is given at \((1, 4)\). At an inflection point, the second derivative \( f''(x) \) is zero but changes sign around \( x \). This means the curve changes from concave up to concave down, or vice versa. You can locate the inflection point by setting the second derivative equal to zero and solving for \( x \). Recognizing inflection points aids in comprehending the overall shape and curvature of a polynomial graph.

Formulating Cubic Polynomial Equations

A cubic polynomial equation is expressed as \( f(x) = ax^3 + bx^2 + cx + d \), where \( a, b, c, \) and \( d \) are coefficients. The leading coefficient \( a \), given as 1 in this exercise, indicates the degree of the polynomial. For our specific problem, the process involves:

• Setting the derivative \( f'(x) \) equal to zero to find critical points.
• Utilizing the second derivative \( f''(x) \) to find inflection points, like \((1, 4)\), by ensuring \( f''(1) = 0 \).
• Determining constants by using known points. For example, substitute \((1, 4)\) into \( f(x) \) to solve for \( d \).

This method allows us to systematically solve for the coefficients in the polynomial, ensuring the shape and properties of the curve meet the requirements for critical and inflection points.