Problem 143 - Calculus Volume 1

Question

Find a quadratic polynomial such that $$f(1)=5, f^{\prime}(1)=3\( and \)f^{\prime \prime}(1)=-6$$

Step-by-Step Solution

Answer

The quadratic polynomial is \( f(x) = -3x^2 + 9x - 1 \).

Step 1: Define the Quadratic Polynomial

Let's express the quadratic polynomial in a general form: \( f(x) = ax^2 + bx + c \). We need to determine the coefficients \(a\), \(b\), and \(c\).

Step 2: Use the Given Condition \(f(1) = 5\)

Substitute \(x = 1\) into the polynomial to use the condition \(f(1) = 5\):\[f(1) = a(1)^2 + b(1) + c = a + b + c = 5.\] This gives us the first equation: \(a + b + c = 5\).

Step 3: Derive the First Derivative

The first derivative of \(f(x) = ax^2 + bx + c\) is \(f'(x) = 2ax + b\).

Step 4: Use the Condition \(f'(1) = 3\)

Substitute \(x = 1\) into \(f'(x)\): \[ f'(1) = 2a(1) + b = 2a + b = 3. \] This gives us the second equation: \(2a + b = 3\).

Step 5: Derive the Second Derivative

The second derivative of \(f(x) = ax^2 + bx + c\) is \(f''(x) = 2a\).

Step 6: Use the Condition \(f''(1) = -6\)

As \(f''(x) = 2a\) and \(f''(1) = -6\), we have \(2a = -6\). Thus, solve for \(a\): \[ 2a = -6 \Rightarrow a = -3 \].

Step 7: Solve the System of Equations

Substitute \(a = -3\) into the equations from Step 2 and Step 4:1. \( -3 + b + c = 5 \) can be simplified to \(b + c = 8\).2. \( 2(-3) + b = 3 \) gives \(-6 + b = 3\), hence \(b = 9\).Finally, substituting \(b = 9\) into \(b + c = 8\) gives \(9 + c = 8\), thus \(c = -1\).

Step 8: Write the Final Polynomial

With \(a = -3\), \(b = 9\), and \(c = -1\), the quadratic polynomial is \( f(x) = -3x^2 + 9x - 1 \).

Key Concepts

Derivatives

Derivatives are fundamental in calculus and help us understand how a function changes at any given point. When dealing with a quadratic polynomial like \( f(x) = ax^2 + bx + c \), we can find the rate of change at each point by calculating its derivatives. The first derivative, \( f'(x) = 2ax + b \), indicates the slope or rate of change of the original polynomial function. It gives us insight into how rapidly or slowly the values of \( f(x) \) are changing as \( x \) varies.
For the provided exercise, it is mentioned that \( f'(1) = 3 \). This tells us that at \( x = 1 \), the slope of the polynomial is 3, meaning the function is growing at this rate right there. The second derivative, \( f''(x) = 2a \), provides information about the curvature or concavity of the polynomial. It helps us understand whether the function is curving upwards or downwards. In our exercise, \( f''(1) = -6 \) shows that at \( x = 1 \), the function curves downward sharply, indicating it is concave down at this point.

Polynomial Coefficients

Polynomial coefficients \(a\), \(b\), and \(c\) define the specific shape and position of a quadratic polynomial. The term \( a \) affects the curvature direction and width, \( b \) adjusts the slope at various points, and \( c \) sets the vertical shift, acting as the y-intercept.
In the context of our problem, these coefficients are determined by using given conditions: \( f(1) = 5 \), \( f'(1) = 3 \), and \( f''(1) = -6 \). The coefficient \( a \) is found first by the second derivative condition to be \( -3 \). This negative sign indicates a downward parabolic curve, consistent with \( f''(x) \) being negative.
Next, the coefficient \( b \) is derived from \( f'(1) = 3 \), ensuring that the slope at \( x = 1 \) matches the defined value. Finally, \( c \) is adjusted to satisfy the function's value at that point. Together, \(a = -3\), \(b = 9\), and \(c = -1\) form the quadratic polynomial that aligns with all conditions given.

System of Equations

A system of equations arises when you have multiple equations that share the same set of variables and you need to find a common solution for them. In algebra and calculus, these systems are essential for determining unknowns like the coefficients in polynomial equations.
Given the conditions \( f(1) = 5 \), \( f'(1) = 3 \), and \( f''(1) = -6 \), we derive three equations for the quadratic polynomial \( f(x) = ax^2 + bx + c \):

• \( a + b + c = 5 \)
• \( 2a + b = 3 \)
• \( 2a = -6 \)

From these equations, we sequentially solve for \(a\), \(b\), and \(c\). We first solve \(2a = -6\) to find \(a = -3\). Substituting \(a = -3\) into the other equations allows us to solve the system systematically: \(b = 9\) from the second equation, and finally \(c = -1\) by using the remaining equation.
This process illustrates how systems of equations help solve multiple interconnected conditions, revealing the specific characteristics of the polynomial needed for the problem.