Problem 3 - Active Calculus

Question

Consider the one-parameter family of functions given by \(p(x)=x^{3}-a x^{2},\) where \(a>0\) a. Sketch a plot of a typical member of the family, using the fact that each is a cubic polynomial with a repeated zero at \(x=0\) and another zero at \(x=a\). b. Find all critical numbers of \(p\). c. Compute \(p^{\prime \prime}\) and find all values for which \(p^{\prime \prime}(x)=0 .\) Hence construct a second derivative sign chart for \(p\). d. Describe how the location of the critical numbers and the inflection point of \(p\) change as \(a\) changes. That is, if the value of \(a\) is increased, what happens to the critical numbers and inflection point?

Step-by-Step Solution

Answer

Critical numbers: \( x = 0 \) and \( x = \frac{2a}{3} \). Inflection point: \( x = \frac{a}{3} \). As \( a \) increases, critical numbers and inflection point move right.

Step 1: Understand the function family

The function given is a one-parameter family of cubic polynomials, \( p(x) = x^3 - ax^2 \), where \( a > 0 \). It has zeros at \( x = 0 \) and \( x = a \). These zeros indicate where the graph will cross the x-axis.

Step 2: Sketch the function

Since \( p(x) \) is a cubic polynomial with repeated zero at \( x = 0 \) and another zero at \( x = a \), the general shape involves crossing the x-axis at these points. A typical sketch would show the function starting from negative infinity, curving upwards through \( x = 0 \), reaching a peak, then dipping back down, and finally curving upwards again through \( x = a \).

Step 3: Find critical numbers

To find the critical numbers, calculate the first derivative, \( p'(x) \). \[ p'(x) = 3x^2 - 2ax \]. Set the first derivative equal to zero: \[ 3x^2 - 2ax = 0 \]. Solving that, we find: \[ x(3x - 2a) = 0 \]. Hence, the critical numbers are \( x = 0 \) and \( x = \frac{2a}{3} \).

Step 4: Compute the second derivative

Calculate the second derivative, \( p''(x) \): \[ p''(x) = 6x - 2a \]. Set the second derivative equal to zero: \[ 6x - 2a = 0 \]. Solving this yields: \[ x = \frac{a}{3} \]. Thus, the inflection point is at \( x = \frac{a}{3} \).

Step 5: Construct the second derivative sign chart

To construct the sign chart, substitute values around the inflection point \( x = \frac{a}{3} \) into \( p''(x) = 6x - 2a \). Evaluate \( p''(x) \) for \( x < \frac{a}{3} \) and \( x > \frac{a}{3} \):\[ \text{For } x < \frac{a}{3}, \, p''(x) = 6x - 2a < 0 \text{ (concave down)} \. \text{For } x > \frac{a}{3}, \, p''(x) = 6x - 2a > 0 \text{ (concave up)} \].

Step 6: Analyze critical numbers and inflection point as \( a \) changes

As \( a \) increases, the critical number \( x = \frac{2a}{3} \) and the inflection point \( x = \frac{a}{3} \) both move to the right on the x-axis. The zero at \( x = a \) also moves to the right, altering the shape and position of the function's peaks and valleys.

Key Concepts

cubic polynomials

A cubic polynomial is a polynomial of degree three, which means its highest power of the variable is three. The general form of a cubic polynomial is: \[ p(x) = ax^3 + bx^2 + cx + d \] In this exercise, we are given a one-parameter family of cubic polynomials, which simplifies to: \[ p(x) = x^3 - ax^2 \] where \( a > 0 \). Key features of cubic polynomials include up to three roots (where the function crosses the x-axis), possible local maximum and minimum points, and an inflection point, where the concavity changes.

critical numbers

Critical numbers are values of \( x \) at which the first derivative of a function is zero or undefined. These points indicate where the slope of the function changes, which can correspond to local maxima, minima, or saddle points. To find the critical numbers of \( p(x) = x^3 - ax^2 \):

• First, compute the first derivative: \[ p'(x) = 3x^2 - 2ax \]
• Set the derivative equal to zero and solve: \[ 3x^2 - 2ax = 0 \]
• Factor to find: \[ x(3x - 2a) = 0 \]
• Thus, the critical numbers are: \( x = 0 \) and \( x = \frac{2a}{3} \)

These values indicate the points at which the function's slope changes and where the function may have local maximum or minimum values.

second derivative test

The second derivative test helps determine the nature of the critical numbers found by analyzing concavity. To apply this test:

• First, compute the second derivative: \[ p''(x) = 6x - 2a \]
• Find where the second derivative is zero: \[ 6x - 2a = 0 \]. Solving for \( x \), we get: \( x = \frac{a}{3} \), the inflection point.

To determine concavity:

• For \( x < \frac{a}{3} \), \( p''(x) = 6x - 2a < 0 \), indicating the function is concave down.
• For \( x > \frac{a}{3} \), \( p''(x) = 6x - 2a > 0 \), indicating the function is concave up.

This test tells us whether each critical number is a local minimum, maximum, or a saddle point. If \( p''(x) \) is positive at a critical number, it indicates a local minimum; if negative, it indicates a local maximum.

inflection points

An inflection point is where a function changes concavity, from concave up to concave down or vice versa. For our function \( p(x) = x^3 - ax^2 \), the inflection point occurs where \( p''(x) = 0 \): \[ p''(x) = 6x - 2a = 0 \] Solving this equation gives: \[ x = \frac{a}{3} \] At \( x = \frac{a}{3} \), we switch from concave down to concave up. As the parameter \( a \) increases, the inflection point shifts to the right. So, as \( a \) grows larger, both critical numbers \( x = 0 \) and \( x = \frac{2a}{3} \) as well as the inflection point \( x = \frac{a}{3} \) move rightward along the x-axis. This alters both the position and shape of the function's graph, as its peaks and valleys also shift.