Problem 21 - Thomas Calculus in SI Units

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x^{5}-5 x^{4}=x^{4}(x-5)$$

Step-by-Step Solution

Answer

Local minimum at \( x = 4 \), inflection points at \( x = 0 \) and \( x = 3 \).

Step 1: Find the derivative

To identify the critical points of the function, we first find its derivative. The function is given by: \[ y = x^{5} - 5x^{4} \]. Take the derivative: \[ y' = \frac{d}{dx}(x^{5} - 5x^{4}) = 5x^{4} - 20x^{3}. \]

Step 2: Solve for critical points

Critical points occur where the derivative is zero or undefined. The derivative is given by: \[ y' = 5x^{4} - 20x^{3} = 5x^{3}(x - 4). \]Set the derivative to zero:\[ 5x^{3}(x - 4) = 0. \]This implies that the critical points are at \( x = 0 \) and \( x = 4 \).

Step 3: Identify extreme points

To identify whether these critical points are local minima or maxima, use the second derivative test. First, find the second derivative: \[ y'' = \frac{d}{dx}(5x^{4} - 20x^{3}) = 20x^{3} - 60x^{2}. \]Evaluate at the critical points:For \( x = 0 \): \( y''(0) = 0 \).For \( x = 4 \): \( y''(4) = (20 \times 4^{3}) - (60 \times 4^{2}) = 320. \)Since \( y''(4) > 0 \), \( x = 4 \) is a local minimum.

Step 4: Determine inflection points

Inflection points occur where the second derivative changes sign. Set the second derivative \( y'' = 20x^{3} - 60x^{2} \) to zero:\[ 20x^{2}(x - 3) = 0. \]This results in \( x = 0 \) and \( x = 3 \). Check intervals around these points to see if \( y'' \) changes sign:For \( x = 0 \) and \( x = 3 \), the concavity changes, indicating inflection points at both.

Step 5: Graph the function

Use the information derived from the previous steps:- The function has a local minimum at \( x = 4 \).- There are inflection points at \( x = 0 \) and \( x = 3 \).Plot these points and observe the curvature of the graph around these locations using a graphing calculator or graphing software to visualize.

Key Concepts

Derivative Analysis

Analyzing the derivative of a function is key to understanding its behavior. The derivative helps us determine the rate at which the function changes. For the function \( y = x^5 - 5x^4 \), the derivative is obtained by applying basic differentiation rules:

• \( y' = 5x^4 - 20x^3 \)

This new expression, the derivative, allows us to find critical points and investigate where the function increases or decreases. Solving \( y' = 0 \), we find the critical points — places where the slope is zero and potential local maxima or minima can occur.

Critical Points

Critical points occur where the derivative of the function is zero or undefined. These points determine where a function might change direction:

• For our function, we set \( 5x^3(x-4) = 0 \).
• Solving gives us the critical points at \( x = 0 \) and \( x = 4 \).

These points are not automatically minima or maxima. We need further investigation, such as using the second derivative, to identify their nature. Critical points are foundational in graphing because they mark potentially significant changes in the graph's trend.

Extreme Values

Extreme values are the highest or lowest points the function reaches. They can be local (only within a small region) or absolute (through the entire domain). To distinguish if our critical points are extreme values, we use the second derivative test:

• First, calculate the second derivative: \( y'' = 20x^3 - 60x^2 \).
• Then evaluate at critical points.
• For \( x = 0 \): Second derivative is 0, indicating a potential inflection point rather than an extreme.
• For \( x = 4 \): \( y''(4) = 320 \), positive, meaning a local minimum.

Identifying these helps sketch an accurate graph and understand the peaks and valleys of the function.

Inflection Points

Inflection points represent where a graph changes concavity, switching from being concave up (like a cup) to concave down (like an umbrella), or vice versa. We determine these by finding where the second derivative changes sign.

• Set \( y'' = 20x^3 - 60x^2 = 0 \).
• This reveals potential inflection points at \( x = 0 \) and \( x = 3 \).

Check intervals around these values to confirm changes in concavity. Both \( x = 0 \) and \( x = 3 \) indeed produce sign changes in the second derivative. Thus, they're true inflection points, offering insight into the "shape" of the graph.