Problem 119 - Thomas Calculus

Question

Find the values of constants \(a, b,\) and \(c\) so that the graph of \(y=a x^{3}+b x^{2}+c x\) has a local maximum at \(x=3,\) local minimum at \(x=-1,\) and inflection point at \((1,11) .\)

Step-by-Step Solution

Answer

The constants are \(a = -1\), \(b = 3\), and \(c = 9\).

Step 1: Set Up the Derivative Equalities

The local maximum at \(x=3\) and the local minimum at \(x=-1\) imply that the first derivative of the function should be zero at these points. The first derivative of the function \(y = ax^3 + bx^2 + cx\) is \(y' = 3ax^2 + 2bx + c\). We set up the equations: \(3a(3)^2 + 2b(3) + c = 0\) and \(3a(-1)^2 + 2b(-1) + c = 0\).

Step 2: Set Up the Inflection Point Condition

An inflection point at \((1, 11)\) implies that the second derivative of the function changes sign there, and the function value at \(x=1\) is 11. The second derivative is \(y'' = 6ax + 2b\). The equation \(6a(1) + 2b = 0\) ensures a changing concavity. Also, \(a(1)^3 + b(1)^2 + c(1) = 11\).

Step 3: Solve the System of Equations

We now solve the system of equations derived from the previous steps:1. \(27a + 6b + c = 0\) from the maximum condition at \(x=3\).2. \(3a - 2b + c = 0\) from the minimum condition at \(x=-1\).3. \(6a + 2b = 0\) from the inflection point condition.4. \(a + b + c = 11\) from the function value at \((1,11)\).From equation 3, \(b = -3a\). Substitute \(b=-3a\) into the other equations.

Step 4: Substitute and Simplify

Substituting \(b = -3a\) into equations 1 and 2:1. \(27a + 6(-3a) + c = 0\), simplifies to \(27a - 18a + c = 0\), so \(9a + c = 0\).2. \(3a - 2(-3a) + c = 0\), simplifies to \(3a + 6a + c = 0\), so \(9a + c = 0\) (consistently the same as equation 1).Equation 4 becomes \(a - 3a + c = 11\), hence \(-2a + c = 11\).

Step 5: Solve for Constants

From equations \(9a + c = 0\) and \(-2a + c = 11\), solve for \(a\) and \(c\):- Substitute \(c = -9a\) into \(-2a + c = 11\) to get \(-2a - 9a = 11\), leading to \(-11a = 11\) and thus \(a = -1\).- Substitute \(a = -1\) into \(c = -9a\) to get \(c = -9(-1) = 9\).- For \(b\), since \(b = -3a\), substitute \(a = -1\) to get \(b = -3(-1) = 3\).

Step 6: Verify the Solution

Verify by substituting \(a = -1\), \(b = 3\), \(c = 9\) into the original conditions:- Plug into \(27a + 6b + c = 0\) to confirm.- Plug into \(3a - 2b + c = 0\) to confirm.- Verify inflection point conditions \(6a + 2b = 0\) and \(a + b + c = 11\). All equations should hold true.

Key Concepts

Local Maximum and Minimum

Understanding local maximum and minimum points is fundamental in calculus problem solving. These points occur where a function reaches its highest or lowest value locally within its immediate surrounding. To locate these points for a function, we assess its first derivative.

• A local maximum at a certain point suggests that the function changes from increasing to decreasing.
• A local minimum indicates a change from decreasing to increasing.
• Both conditions require the derivative, or the slope of the tangent line, to be zero at those points.

For the function given, the local maximum at \( x = 3 \) and the local minimum at \( x = -1 \) imply that the derivative \( y' = 3ax^2 + 2bx + c \) becomes zero at these points.

Inflection Point

An inflection point occurs when a curve changes its concavity, switching from being concave upwards to concave downwards or vice versa. For this exercise, at the inflection point of \((1, 11)\), both the second derivative and the function value conditions need to be satisfied:

• The second derivative \( y'' = 6ax + 2b \) changes its sign at \( x = 1 \).
• Additionally, the function \( y = ax^3 + bx^2 + cx \) assumes a value of 11 when \( x = 1 \).

Meeting these criteria involves ensuring that the equation \( 6a + 2b = 0 \) is satisfied and aligning the function value with the given point by solving \( a + b + c = 11 \). This ensures the curve not only changes its shape but aligns with the specified point on the graph.

System of Equations

The utilization of a system of equations is paramount when solving for multiple unknowns, such as the constants \(a, b, \) and \(c\) in this function. The step-by-step solution involves setting up and solving these equations:

• From the local maximum \(27a + 6b + c = 0\).
• From the local minimum \(3a - 2b + c = 0\).
• The inflection point leads to \(6a + 2b = 0\).
• The point value requires \(a + b + c = 11\).

By substituting one equation into another, we find the values of \(a, b,\) and \( c \) that satisfy all conditions. This often includes simplifying and reducing the system by expressing one variable in terms of another, such as \( b = -3a \) from the second derivative condition.

Derivative Analysis

Derivative analysis forms the backbone of calculus and helps in understanding the behavior of functions. By examining the first and second derivatives, one can gather comprehensive information about a function's graph, including the local extremes and points of inflection.

• The first derivative \( y' = 3ax^2 + 2bx + c \) is critical for identifying where the slope of the curve is zero and thus identifying local maxima and minima.
• The second derivative \( y'' = 6ax + 2b \) helps determine where the curvature changes, indicating inflection points.

The process includes setting derivative equations equal to zero and using these conditions to solve for unknowns in the function. Such analysis not only identifies key features of the graph but also sheds light on the function's overall behavior.