Problem 20 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=x^{4}+2 x^{3}=x^{3}(x+2) \end{equation}

Step-by-Step Solution

Answer

Critical points are at \( x = 0 \) and \( x = -\frac{3}{2} \). Inflection points at \( x = 0 \) and \( x = -\frac{3}{2} \).

Step 1: Find the first derivative

To find the local extreme points, we need the first derivative of the function. Start with the original function: \[ y = x^4 + 2x^3 = x^3(x+2) \]. Differentiate this function with respect to \( x \) using the product rule. The derivative is: \[ y' = x^3 rac{d}{dx}(x+2) + (x+2) rac{d}{dx}(x^3) = x^3 imes 1 + 3x^2 imes (x+2) \]. Simplify it to: \[ y' = x^3 + 3x^2(x+2) = x^3 + 3x^3 + 6x^2 = 4x^3 + 6x^2 \]. The first derivative is \( y' = 2x^2(2x + 3) \).

Step 2: Find critical points

Set the first derivative equal to zero to find the critical points: \[ y' = 2x^2(2x+3) = 0 \].This equation is satisfied when \( 2x^2 = 0 \) or \( 2x+3 = 0 \). First, solve \( 2x^2 = 0 \) which gives \( x = 0 \). Then, solve \( 2x+3 = 0 \) which gives \( x = -\frac{3}{2} \). Thus, the critical points are \( x = 0 \) and \( x = -\frac{3}{2} \).

Step 3: Determine local extrema

To determine whether these critical points are maxima or minima, use the second derivative test. First find the second derivative \( y'' \) from \( y' = 2x^2(2x+3) \). Apply the product rule and simplify:\[ y'' = d/dx [2x^2(2x+3)] = 2*(4x^2 + 6x) = 8x^2 + 12x \].Evaluate \( y'' \) at \( x = 0 \) and \( x = -\frac{3}{2} \) to classify:- \( y''(0) = 0 \), requires more analysis.- \( y''(-\frac{3}{2}) = 8(-\frac{3}{2})^2 + 12(-\frac{3}{2}) = 18 - 18 = 0 \), inconclusive by the second derivative test.Thus, use first derivative tests or other methods if required for behavior analysis at critical points.

Step 4: Find the second derivative

Continue analyzing for inflection points. Use the same second derivative, \( y'' = 8x^2 + 12x \). Set this equal to zero to find possible inflection points: \[ 8x^2 + 12x = 0 \].Factor the equation: \[ 4x(2x+3) = 0 \].This gives solutions \( x=0 \) and \( x=-\frac{3}{2} \). Inflection points occur where the concavity changes, so test intervals around these points.

Step 5: Graph analysis

Once critical points and inflection points are found, graph the function for visual confirmation. Plot the function: \( y = x^3(x+2) \).- Critical points: \( x = 0 \) and \( x = -\frac{3}{2} \).- Inflection points: \( x = 0 \) and \( x = -\frac{3}{2} \); check if any additional concavity points appear by evaluating the second derivative near these points.Graphing will provide a clear visual of the extrema and inflection behavior.

Key Concepts

Critical Points

Critical points are vital in understanding where a function changes its behavior, particularly from increasing to decreasing or vice versa. They are found by setting the first derivative of the function equal to zero and solving for the variable.
- In our exercise, the function is given by: \[ y = x^3(x+2) \].
- The first derivative, denoted as \( y' \), is calculated and set to zero: \[ y' = 2x^2(2x+3) = 0 \].
This results in critical points at \( x = 0 \) and \( x = -\frac{3}{2} \).
At these points, the function's slope is zero, indicating potential locations for local minima or maxima or possibly points of inflection.
Determining the nature of these points involves further tests, such as evaluating changes in the sign of the derivative or using the second derivative test.

Derivative

The derivative of a function is a fundamental concept in calculus that tells us how the function changes as its input changes. It represents the slope of the tangent line to the function at any point on its curve, giving insight into whether the function is increasing or decreasing.
- For the function \( y = x^3(x+2) \), the first derivative \( y' \) is found using the product rule: \[ y' = 4x^3 + 6x^2 \].
- Simplifying further yields: \[ y' = 2x^2(2x + 3) \].
This expression shows that the rate of change of the function is dependent on the inputs \( x \) and is crucial for locating critical points and understanding the function's behavior.
Finding and using derivatives is essential for analysis of local extrema, as discussed in the critical points section.

Inflection Points

Inflection points are where a function changes its concavity, switching from concave up to concave down, or vice versa. Identifying inflection points involves analyzing the second derivative of the function.
- For our function, the second derivative \( y'' \) is given by: \[ y'' = 8x^2 + 12x \].
- Setting \( y'' \) equal to zero gives potential inflection points: \[ 8x^2 + 12x = 0 \].
- Factoring leads to solutions \( x = 0 \) and \( x = -\frac{3}{2} \).
An inflection point is confirmed if the concavity changes across these points. This means checking the sign of \( y'' \) before and after these \( x \)-values to ensure a switch from positive to negative or vice versa.
Recognizing such points on a graph helps in understanding the curvature and behavior of the entire function.

Second Derivative Test

The second derivative test is a method used to classify the critical points found by the first derivative as relative maxima, minima, or saddle points.
- To perform the test, we use the second derivative \( y'' \) of our function: \[ y'' = 8x^2 + 12x \].
- Evaluate \( y'' \) at each critical point:

• At \( x = 0 \), \( y''(0) = 0 \), which is inconclusive and requires further analysis.
• At \( x = -\frac{3}{2} \), \( y''(-\frac{3}{2}) = 0 \) also proves inconclusive.

In situations where the second derivative test does not clearly indicate maxima or minima, alternative methods such as the first derivative test or analyzing the function's behavior must be employed.
This test typically assists in making quick determinations once the concavity is confirmed, aiding in the broader analysis of the function's graph.