Problem 49 - Calculus A New Horizon

Question

(a) Prove that a general cubic polynomial $$f(x)=a x^{3}+b x^{2}+c x+d \quad(a \neq 0)$$has exactly one inflection point. (b) Prove that if a cubic polynomial has three \(x\) -intercepts. then the inflection point occurs at the average value of the intercepts. (c) Use the result in part ( \(b\) ) to find the inflection point of the cubic polynomial \(f(x)=x^{3}-3 x^{2}+2 x,\) and check your result by using \(f^{\prime \prime}\) to determine where \(f\) is concave up and concave down.

Step-by-Step Solution

Answer

(a) Inflection point at \( x = -\frac{b}{3a} \); (b) occurs at the average of x-intercepts; (c) inflection point at \( x = 1 \).

Step 1: Finding the Second Derivative of f(x)

The second derivative of a cubic polynomial shows where the rate of change of the slope, or concavity, changes. Given \( f(x) = ax^3 + bx^2 + cx + d \), the first derivative is \( f'(x) = 3ax^2 + 2bx + c \). The second derivative is \( f''(x) = 6ax + 2b \).

Step 2: Finding the Inflection Point

An inflection point occurs where the second derivative is equal to zero. Solving \( 6ax + 2b = 0 \) gives \( x = -\frac{b}{3a} \). This shows that the cubic function has exactly one inflection point.

Step 3: Inflection Point for Three X-Intercepts

For a cubic polynomial \( f(x) = a(x-r_1)(x-r_2)(x-r_3) \), the x-intercepts are \( r_1, r_2, \) and \( r_3 \). The average x-intercept is \( \frac{r_1 + r_2 + r_3}{3} \). The derived location of the inflection point in Step 2, \( x = -\frac{b}{3a} \), is the same as this average if the polynomial is expressed in its factored form.

Step 4: Apply to Specific Polynomial f(x) = x^3 - 3x^2 + 2x

For \( f(x) = x^3 - 3x^2 + 2x \), identify that the roots are \( x = 0, 1, \) and \( 2 \). The average of these roots is \( \frac{0 + 1 + 2}{3} = 1 \).

Step 5: Verify Using Second Derivative

For \( f(x) = x^3 - 3x^2 + 2x \), the second derivative is \( f''(x) = 6x - 6 \). Set \( f''(x) = 0 \) to find the inflection point, solving \( 6x - 6 = 0 \) gives \( x = 1 \). This confirms the result from Step 4.

Step 6: Analyzing Concavity Using f''(x)

At \( x = 1 \), \( f''(x) = 0 \). For \( x < 1 \), let \( x = 0 \), \( f''(0) = -6 \) implies concave down. For \( x > 1 \), let \( x = 2 \), \( f''(2) = 6 \) implies concave up.

Key Concepts

Second Derivative

The second derivative of a function helps us understand how the rate of change of the slope behaves. For a cubic polynomial given by \( f(x) = ax^3 + bx^2 + cx + d \), we begin by finding its derivatives.
The first derivative, \( f'(x) = 3ax^2 + 2bx + c \), indicates the slope of the tangent line at each point along the curve.
Next, we find the second derivative, which is \( f''(x) = 6ax + 2b \).

• This second derivative gives us a crucial insight into where the graph of the function might change its curvature from concave up to concave down or vice versa.
• At these points where \( f''(x) = 0 \), the concavity changes, indicating a potential inflection point.

Inflection Point

An inflection point is where the graph of a function changes its concavity. For a function to have an inflection point:

• The second derivative, \( f''(x) \), must be zero or undefined at that point.
• This does not automatically mean there is an inflection point, but it is a prerequisite for finding one.

For our cubic polynomial \( f(x) = ax^3 + bx^2 + cx + d \), solving \( f''(x) = 6ax + 2b = 0 \) gives the inflection point at \( x = -\frac{b}{3a} \).
This result shows that any cubic function has exactly one inflection point, because the second derivative is linear, crossing the x-axis at most once.

Concavity

Concavity tells us whether a function is bending upwards or downwards at a particular section of its curve. This concept is essential in analyzing the shape of graph:

• If the second derivative \( f''(x) > 0 \), the function is concave up at that interval, resembling a U-shape.
• If \( f''(x) < 0 \), the function is concave down at that interval, resembling an upside-down U.

By analyzing \( f(x) = x^3 - 3x^2 + 2x \):

• We calculate the second derivative as \( f''(x) = 6x - 6 \).
• At an inflection point, \( x = 1 \), \( f''(1) = 0 \), indicating a change in concavity.
• Checking to the left \( x = 0 \), the second derivative is \(-6\), meaning concave down.
• Checking to the right \( x = 2 \), the second derivative is \(6\), meaning concave up.

X-Intercepts

The x-intercepts or roots of the function are the points where the graph of a polynomial crosses the x-axis. These are found by setting the function equal to zero and solving for x.

• For a cubic formula expressed as \( f(x) = a(x-r_1)(x-r_2)(x-r_3) \), the roots are \( r_1, r_2, r_3 \).
• Each root represents an x-intercept, where the function has values of zero.

For \( f(x) = x^3 - 3x^2 + 2x \), the roots are found at \( x = 0, 1, \) and \( 2 \). These roots correspond to where the graph intersects the x-axis, showcasing how the function's value changes sign across these points.

Average Value of Intercepts

Finding the average value of x-intercepts provides valuable insight into the geometry of the cubic polynomial's curve. It is especially pivotal in relating to the inflection point:

• The average value of intercepts is calculated as \( \frac{r_1 + r_2 + r_3}{3} \).
• In the context of cubic polynomials, this average value aligns with the x-coordinate of the inflection point found via the second derivative.

In our example, with roots \( x = 0, 1, \) and \( 2 \), the average intercept is \( \frac{0 + 1 + 2}{3} = 1 \).
This corresponds precisely to the inflection point \( x = 1 \), reinforcing our earlier calculations regarding the change in concavity.