Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-18, use the Concavity Theorem to determine where the given function is concave up and where it is concave down. Also find all inflection points. \(q(x)=x^{4}-6 x^{3}-24 x^{2}+3 x+1\)

Step-by-Step Solution

Answer

The function is concave up on \((-\infty, -1)\) and \((4, \infty)\) and concave down on \((-1, 4)\). Inflection points are at \((-1, -19)\) and \((4, -499)\).

Step 1: Find the First Derivative

First, we need to find the first derivative of the function, \( q(x) = x^4 - 6x^3 - 24x^2 + 3x + 1 \). We differentiate term by term: \[ q'(x) = 4x^3 - 18x^2 - 48x + 3. \] This derivative will be used to find critical points and analyze intervals.

Step 2: Find the Second Derivative

Next, we find the second derivative of the function, \( q'(x) = 4x^3 - 18x^2 - 48x + 3 \). Again, differentiate term by term: \[ q''(x) = 12x^2 - 36x - 48. \] The second derivative will help us determine the concavity and locate inflection points.

Step 3: Solve \( q''(x) = 0 \) for Possible Inflection Points

Set the second derivative equal to zero to find where the concavity may change: \( 12x^2 - 36x - 48 = 0 \). We solve for \( x \) using the quadratic formula or by simplifying the expression. Factoring gives: \( 12(x^2 - 3x - 4) = 0 \) or \( (x - 4)(x + 1) = 0 \). Thus, potential inflection points are \( x = 4 \) and \( x = -1 \).

Step 4: Determine Concavity on Intervals

Use \( x = 4 \) and \( x = -1 \) to divide the number line into intervals. Test each interval in the second derivative \( q''(x) \):- For \( x < -1 \), choose \( x = -2 \), \( q''(-2) = 12(4) + 72 - 48 = 72 \) (concave up).- For \( x \in (-1, 4) \), choose \( x = 0 \), \( q''(0) = -48 \) (concave down).- For \( x > 4 \), choose \( x = 5 \), \( q''(5) = 12(25) - 180 - 48 = 72 \) (concave up).

Step 5: Identify Inflection Points

Based on the change in sign of \( q''(x) \), the concavity changes at \( x = -1 \) and \( x = 4 \). Thus, these are inflection points. To determine the coordinate of these points, substitute back into the original function:- At \( x = -1 \), \( q(-1) = 1 + 6 - 24 - 3 + 1 = -19 \), so \((-1, -19)\) is an inflection point.- At \( x = 4 \), \( q(4) = 256 - 384 - 384 + 12 + 1 = -499 \), so \((4, -499)\) is another inflection point.

Key Concepts

Inflection Points

Inflection points represent key positions on a graph where the curvature changes direction. They are identified where the second derivative of a function equals zero and the concavity changes from being upward to downward, or vice versa.
To discover inflection points:

• Compute the second derivative of the function.
• Set the second derivative equal to zero to find potential inflection points.
• Check intervals around these points to verify where the sign of the second derivative changes.

In the example with the function \( q(x) = x^4 - 6x^3 - 24x^2 + 3x + 1 \), we found the potential inflection points at \( x = -1 \) and \( x = 4 \) by solving the equation \( q''(x) = 0 \). The sign change around these values confirmed them as true inflection points. At \( x = -1 \) and \( x = 4 \), concavity changes, indicating these positions as inflection points on the curve.

First Derivative

The first derivative of a function reflects its rate of change, akin to a function's slope at any given point. It is fundamental in discovering where a function has critical points, which often lead to maximas or minimas, but also serve as key information for further analysis of concavity.
For the function \( q(x) \), the first derivative is calculated as \( q'(x) = 4x^3 - 18x^2 - 48x + 3 \). This expression will not directly indicate concavity, but it is crucial in breaking down the function for deeper analysis.
Additionally, when the first derivative changes sign, it points out critical points which can sometimes be used as a reference to further validate inflection points discovered with the second derivative.

Second Derivative

The second derivative of a function, often symbolized as \( q''(x) \), provides insight into the function's concavity. It determines whether the graph of a function is concave up or concave down at a given interval.
For our function \( q(x) \), the second derivative is \( q''(x) = 12x^2 - 36x - 48 \). This derivative plays a crucial role in determining where the graph curves upwards or downwards.

• When \( q''(x) > 0 \), the function is concave up.
• When \( q''(x) < 0 \), the function is concave down.

Solving \( q''(x) = 0 \) provided us with potential transition points, and evaluation around these points gave insights into concavity changes, confirming inflection points.

Concavity Analysis

Concavity analysis allows us to understand the "bending" behavior of a graph. By examining the intervals of concavity, we gain insight into the general shape and behavior of the function.
To perform a concavity analysis:

• Divide the number line into intervals based on potential inflection points (or roots of the second derivative).
• Test each interval using the second derivative \( q''(x) \) to determine its sign.
• A positive sign indicates concavity upwards, while a negative sign reveals concavity downwards.

In this exercise, by checking the intervals around \( x = -1 \) and \( x = 4 \) with test points, we deduced:

• For \( x < -1 \), the function is concave up.
• For \( x \in (-1, 4) \), the function is concave down.
• For \( x > 4 \), the function is concave up again.

This comprehensive analysis is central to understanding how a function behaves across different intervals and is pivotal in graphing the function effectively.