Problem 36 - Thomas Calculus

Question

What values of \(a\) and \(b\) make \(f(x)=x^{3}+a x^{2}+b x\) have \begin{equation}\begin{array}{l}{\text { a. a local maximum at } x=-1 \text { and a local minimum at } x=3 ?} \\ {\text { b. a local minimum at } x=4 \text { and a point of inflection at } x=1 ?}\end{array}\end{equation}

Step-by-Step Solution

Answer

Part A: \(a = 3, b = -6\). Part B: \(a = -3, b = -24\).

Step 1: Determine Critical Points for Part A

For the function to have a local maximum at \(x = -1\) and a local minimum at \(x = 3\), its first derivative \(f'(x)\) must equal zero at these points. Start by finding the derivative of \(f(x) = x^3 + ax^2 + bx\). This is \(f'(x) = 3x^2 + 2ax + b\). Set \(f'(-1)\) and \(f'(3)\) to zero to find equations for \(a\) and \(b\).

Step 2: Solve for Part A

Solve the equations \(3(-1)^2 + 2a(-1) + b = 0\) and \(3(3)^2 + 2a(3) + b = 0\). Simplifying gives two linear equations: \(3 - 2a + b = 0\) and \(27 + 6a + b = 0\). Solve this system of equations to find the values of \(a\) and \(b\).

Step 3: Determine Critical Points and Inflection for Part B

To have a local minimum at \(x = 4\) and a point of inflection at \(x = 1\), the first derivative \(f'(x)\) must be zero at \(x = 4\). Use the second derivative \(f''(x)\) to evaluate inflection by finding where \(f''(x)\) changes sign. Find \(f''(x) = 6x + 2a\) and ensure it is zero at \(x = 1\) for an inflection point.

Step 4: Solve for Part B

Set the equations \(3(4)^2 + 2a(4) + b = 0\) from \(f'(4) = 0\) and \(6(1) + 2a = 0\) from \(f''(1) = 0\). Solving \(48 + 8a + b = 0\) and \(2a = -6\) gives \(a = -3\). Substitute \(a = -3\) into the first equation from this part and solve for \(b\) to finish part b.

Key Concepts

Critical Points

Critical points are where the first derivative of a function is zero or undefined. These points are essential when analyzing the behavior of functions, as they can indicate where the function might have local maximums, minimums, or points of inflection. In calculus, to find the critical points:

• Take the first derivative of the function, which we'll denote as \( f'(x) \).
• Set \( f'(x) = 0 \) to solve for the values of \( x \) where the derivative equals zero. These are your critical points.
• Also check where \( f'(x) \) is undefined, though for most polynomial functions, it's generally defined everywhere.

In our case with \( f(x) = x^3 + ax^2 + bx \), the derivative is \( f'(x) = 3x^2 + 2ax + b \). We look for critical points by setting this equation to zero.

Local Maximum and Minimum

Determining local maximums and minimums forms a crucial part of optimization problems in calculus. A local maximum is a point where a function reaches a peak over a small interval, while a local minimum is where it hits a trough.To find these points:

• First, identify the critical points using the derivative as explained in the previous section.
• Use the second derivative test: calculate the second derivative of the function, \( f''(x) \).
• Plug the critical points into \( f''(x) \):
  • If \( f''(x) > 0 \), the function has a local minimum at that point because the graph is concave up.
  • If \( f''(x) < 0 \), there is a local maximum because the graph is concave down.

For example, the exercise specifies finding \( a \) and \( b \) such that the function \( f(x) \) has a local maximum at \( x = -1 \) and a local minimum at \( x = 3 \). This requires solving the derivatives for these points and using the second derivative test to confirm the nature of these critical points.

Point of Inflection

A point of inflection is where a function changes concavity, which means it shifts from being concave up to concave down, or vice versa. Identifying these points involves the second derivative of the function.To find a point of inflection:

• Find the second derivative of the function, noted as \( f''(x) \).
• Set \( f''(x) = 0 \) and solve for \( x \). These points are potential inflection points.
• Check if the sign of \( f''(x) \) changes around these potential points to confirm an actual change in concavity.

In part b of the exercise, the point of inflection at \( x = 1 \) requires that \( f''(x) = 0 \). Using \( f''(x) = 6x + 2a \) for our function, setting it zero at \( x = 1 \) helps us solve for a critical parameter in the exercise.

Derivatives

Understanding derivatives is vital as they provide information about the rate of change in a function. First derivatives, like \( f'(x) \), tell us about the function's slope at any point \( x \). They are the foundational tools for finding critical points, determining the behavior of a function (like increasing or decreasing), and identifying local extremes.The second derivative, \( f''(x) \), provides insight into the concavity of the function:

• If \( f''(x) > 0 \), the function is concave up, resembling a cup that holds water.
• If \( f''(x) < 0 \), it's concave down, like an upside-down cup.

In our original exercise, we used the first derivative \( f'(x) = 3x^2 + 2ax + b \) to locate critical points and the second derivative \( f''(x) = 6x + 2a \) to determine concavity and points of inflection. These derivatives are crucial for understanding the full behavior of polynomial functions.