Problem 30 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

For part (a): \(a = -3\), \(b = -9\); for part (b): \(a = -3\), \(b = -24\).

Step 1: Calculating the first derivative

First, we need to find the first derivative of the function \( f(x) = x^3 + ax^2 + bx \). The first derivative, \( f'(x) \), will help identify critical points where the local maxima and minima occur.\[ f'(x) = \frac{d}{dx}(x^3 + ax^2 + bx) = 3x^2 + 2ax + b \]

Step 2: Critical Point Equations for Part (a)

For part (a), we set the derivative equal to zero to find the local maxima and minima:\[ 3x^2 + 2ax + b = 0 \]Since we are given that there's a local maximum at \( x = -1 \) and a local minimum at \( x = 3 \), substituting these, we get two equations:1. \( 3(-1)^2 + 2a(-1) + b = 0 \) simplifies to: \( 3 - 2a + b = 0 \).2. \( 3(3)^2 + 2a(3) + b = 0 \) simplifies to: \( 27 + 6a + b = 0 \).

Step 3: Solving the Equations for Part (a)

We'll now solve the system of equations obtained:1. \( 3 - 2a + b = 0 \)2. \( 27 + 6a + b = 0 \)Subtract the first equation from the second to eliminate \( b \):\[ (27 + 6a + b) - (3 - 2a + b) = 0 \]This yields:\[ 24 + 8a = 0 \]Solving for \( a \), we find:\[ a = -3 \]Substitute \( a = -3 \) back into either equation (say the first):\[ 3 - 2(-3) + b = 0 \]\[ 3 + 6 + b = 0 \]\[ b = -9 \]

Step 4: Critical Point and Point of Inflection Equations for Part (b)

For part (b), we'll still use the derivative for critical points and point of inflection:Set the first derivative to zero for the minimum at \( x = 4 \):\[ 3(4)^2 + 2a(4) + b = 0 \]Simplifies to:\[ 48 + 8a + b = 0 \]For the point of inflection at \( x = 1 \), we need the second derivative:\[ f''(x) = \frac{d}{dx}(3x^2 + 2ax + b) = 6x + 2a \]At point of inflection, \( f''(1) = 0 \):\[ 6(1) + 2a = 0 \]\[ 6 + 2a = 0 \]

Step 5: Solving the Equations for Part (b)

From the point of inflection equation, solve for \( a \):\[ 2a = -6 \]\[ a = -3 \]Substitute \( a = -3 \) back into the critical point equation:\[ 48 + 8(-3) + b = 0 \]\[ 48 - 24 + b = 0 \]\[ b = -24 \]

Key Concepts

Critical Points

In calculus, critical points are essential for understanding where functions reach their highest or lowest points locally. A critical point occurs on the graph of a function when the derivative is zero or undefined, which might indicate potential local maxima or minima. To find these points, you calculate the first derivative, set it equal to zero, and solve for the values of the variable. In our given problem, the function is:

• \( f(x) = x^3 + ax^2 + bx \)

The first derivative is crucial because:

• \( f'(x) = 3x^2 + 2ax + b \)

Setting \( f'(x) = 0 \) helps identify the values of \(x\) where critical points might occur. Here, this helps us locate where the local maximum and minimum take place as described in the exercise.

Local Maximum

A local maximum is a point where a function changes direction from increasing to decreasing, leading to a peak in the graph. When a function has a local maximum at a certain point, the slope of the tangent at that point is zero, indicating a neutral slope before it shifts downward.

For the given function, we need to ensure a local maximum at \(x = -1\). This implies that when we plug \(x = -1\) into the derivative function:

• \( f'(-1) = 3(-1)^2 + 2a(-1) + b = 0 \)

The zero value of the derivative, along with further analysis of the second derivative, confirms that a peak indeed exists at \(x = -1\) for the predetermined values of \(a\) and \(b\). Calculating these parameters correctly (as shown in the exercise) ensures we capture the local maximum accurately.

Local Minimum

A local minimum refers to a point where a function's graph bottoms out before rising again. At this juncture in the curve, the function's derivative equals zero, marking the lowest point in its immediate vicinity.

Considering our function, to establish a local minimum at \(x = 3\) (in part a of the original exercise), this section of the curve must have:

• \( f'(3) = 3(3)^2 + 2a(3) + b = 0 \)

This ensures the curve's direction reverses. A point is a local minimum if the second derivative is positive there. Solving these equations lets us find \(a\) and \(b\) to align with these local behaviors.

Point of Inflection

A point of inflection is a spot where the curve on a graph changes its concavity. This means where it switches from being concave up (curving upwards) to concave down (curving downwards), or vice versa. Unlike maxima or minima, points of inflection are found by looking at the second derivative.

For the function given, we calculate the second derivative to find such points:

• \( f''(x) = 6x + 2a \)

When you're at a point of inflection, the second derivative is zero:

• \( f''(1) = 0 \)

The second derivative zeroing at \(x = 1\) implies a change in concavity at this point if the concavity itself swaps signs. Therefore, the point of inflection \(x = 1\) leads us to deduce important realities about \(a\) and \(b\) in the original equation.