Problem 12 - The Calculus with Analytic Geometry

Question

If \(f(x)=a x^{3}+b x^{2}+c x\), determine \(a, b\), and \(c\) so that the graph of \(f\) will have a point of inflection at \((1,2)\) and so that the slope of the inflectional tangent there will be \(-2\).

Step-by-Step Solution

Answer

The values are \( a = 4 \), \( b = -12 \), and \( c = 10 \).

Step 1: Define the function and its derivatives

Given the function \( f(x) = a x^{3} + b x^{2} + c x \), find the first and second derivatives; \[ f'(x) = 3a x^2 + 2b x + c \] \[ f''(x) = 6a x + 2b \]

Step 2: Condition for Point of Inflection

A point of inflection occurs where the second derivative is zero and changes sign. At \( x = 1 \), set \( f''(1) = 0 \): \[ 6a (1) + 2b = 0 \] Simplifying, \[ 6a + 2b = 0 \] which gives \[ 3a + b = 0 \] (Equation 1)

Step 3: Use the Inflection Point Coordinates

Use the coordinates of the point of inflection \( (1, 2) \): \[ f(1) = 2 \] Substituting \( x = 1 \) into the function, we get: \[ a(1)^3 + b(1)^2 + c(1) = 2 \] Simplifying, \[ a + b + c = 2 \] (Equation 2)

Step 4: Condition for Slope at Inflection Point

The slope of the tangent at the point of inflection \( (1, 2) \) is given as \(-2\). Therefore, \[ f'(1) = -2 \] Substituting \( x = 1 \) into the first derivative, we get: \[ 3a(1)^2 + 2b(1) + c = -2 \] Simplifying, \[ 3a + 2b + c = -2 \] (Equation 3)

Step 5: Solve the System of Equations

Solve the system of linear equations derived from the conditions: 1. \( 3a + b = 0 \) 2. \( a + b + c = 2 \) 3. \( 3a + 2b + c = -2 \) Step by step: - From Equation 1, we have \( b = -3a \). - Substitute \( b \) in Equation 2, \[ a - 3a + c = 2 \] \[ -2a + c = 2 \] which gives \[ c = 2a + 2 \] Substitute \( b \) and \( c \) into Equation 3, \[ 3a + 2(-3a) + (2a + 2) = -2 \] Simplify, \[ 3a - 6a + 2a + 2 = -2 \] \[ -a + 2 = -2 \] \[ -a = -4 \] \[ a = 4 \] - Then, \( b = -3(4) = -12 \). - Finally, \( c = 2(4) + 2 = 10 \).

Key Concepts

calculus differentiation

Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which represents the rate at which the function's value changes. Given a function like f(x) = a x^{3} + b x^{2} + c x , we find its first derivative to understand how the function behaves.

Steps to differentiate:

• For each term in the function, multiply the coefficient by the power of x.
• Then, decrease the power of x by one.

For example, the first derivative ofa x^{3} + b x^{2} + c x is found as follows:

The derivative of \( ax^{3} \) is \( 3ax^{2} \) since multiplying the coefficient by the power of x (a * 3) gives you \(3a\), and reducing the power by one results in \(x^2\).
Similarly, the derivative of \( bx^{2} \) is \( 2bx\), and the derivative of \( cx\) is simply \(c\).
Therefore, f'(x) = 3a x^2 + 2b x + c is the first derivative (also called the slope).
Differentiation helps in analyzing the slope and curves of functions, which is crucial for graphing and understanding the behavior of functions.

second derivative

The second derivative provides deeper insight into the function's curvature and concavity. It is simply the derivative of the first derivative. Let's continue with our function f(x) = a x^{3} + b x^{2} + c x .

First, we find the second derivative from the first derivative given by 3a x^{2} + 2b x + c . The second derivative is:
f''(x) = 6a x + 2b .
The second derivative tells us about the concavity of the graph:

• A positive second derivative indicates the graph is concave up (shaped like a U).
• A negative second derivative indicates the graph is concave down (shaped like an n).

The second derivative throughout the function helps identify points of inflection, where the curve changes its concavity. At these points, f''(x) = 0 . For our exercise, the point of inflection is at x = 1 with f(1) = 2 and we solve 6a + 2b = 0 to find specific values of a, b, and c ensuring this behavior.

tangent slope

The slope of a tangent line to a curve at a point represents the instantaneous rate of change of the function at that point. It is given by the first derivative. For the specific problem, the slope of the tangent at the inflection point (1, 2) is -2 .

We use the first derivative f'(x)= 3a x^2 + 2b x + c and plug in x = 1 . Setting f'(1) = -2 gives us:
3a(1)^2 + 2b(1) + c = -2
Simplifying this, we obtain a system of equations to determine a, b, and c .

Understanding tangent slopes is crucial:

• A positive slope means the function is increasing at that point.
• A negative slope means it is decreasing.
• A zero slope (horizontal tangent) indicates a local maxima or minima.

For the exercise, after solving the system of equations, we find the values a = 4, b = -12, and c = 10 ensuring that our conditions for the inflection point and the specified tangent slope are satisfied.