Problem 17 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x^{4}-2 x^{2}=x^{2}\left(x^{2}-2\right)$$

Step-by-Step Solution

Answer

Local max at (0, 0) and local mins at (±1, -1). Inflection points at approximately ±0.577.

Step 1: Identify Critical Points

To find critical points, we first need to find the derivative of the function.Given:\[ y = x^4 - 2x^2 \]Take the derivative:\[ y' = 4x^3 - 4x = 4x(x^2 - 1) \]Set the derivative equal to zero to find critical points:\[ 4x(x^2 - 1) = 0 \]This gives us critical points at \( x = 0 \) and \( x = \pm 1 \).

Step 2: Determine Local Extrema

To determine whether the critical points are maxima or minima, evaluate the second derivative:\[ y'' = (4x^3 - 4x)' = 12x^2 - 4 \]Evaluate the second derivative at the critical points:- At \( x = 0 \): \( y''(0) = 12(0)^2 - 4 = -4 \) (concave down, so local max)- At \( x = \pm 1 \): - \( y''(1) = 12(1)^2 - 4 = 8 \) (concave up, so local min) - \( y''(-1) = 12(-1)^2 - 4 = 8 \) (concave up, also local min)

Step 3: Identify Inflection Points

An inflection point occurs where the second derivative changes sign. Solving \( y'' = 0 \) gives:\[ 12x^2 - 4 = 0 \]\[ 12x^2 = 4 \]\[ x^2 = \frac{1}{3} \]\[ x = \pm \sqrt{\frac{1}{3}} \]At these points, check if the second derivative sign changes:- Between \( x = \sqrt{\frac{1}{3}} \text{ and } -\sqrt{\frac{1}{3}} \), verify \( y'' \) changes from negative to positive or vice versa, confirming inflection points.

Step 4: Find Function Values

Calculate the function values at critical and inflection points.- \( y(0) = 0^4 - 2 \times 0^2 = 0 \)- \( y(1) = 1^4 - 2 \times 1^2 = -1 \)- \( y(-1) = (-1)^4 - 2 \times (-1)^2 = -1 \)- \( y(\sqrt{\frac{1}{3}}) \text{ and } y(-\sqrt{\frac{1}{3}}) \) need calculating based on substitution.

Step 5: Graph the Function

To graph the function, plot the key points: local maximum at \( (0, 0) \), local minima at \( (1, -1) \) and \( (-1, -1) \), and inflection points near \( x \approx \pm 0.577 \). The function is symmetric with respect to the y-axis.

Key Concepts

Critical Points

Critical points of a function are where the derivative is either zero or undefined. These points are significant as they help locate where a function could have a maximum or minimum value.
Consider the function given in the exercise:

• To find critical points, calculate the first derivative.
• Set this derivative equal to zero.
• For the function \( y = x^4 - 2x^2 \), the derivative \( y' = 4x^3 - 4x \) simplifies to \( y' = 4x(x^2 - 1) \).
• This results in critical points at \( x = 0 \), \( x = 1 \), and \( x = -1 \).

These points tell us where the slope of the tangent is zero, indicating potential extremes.

Local Extrema

Local extrema refer to the highest or lowest points in a specific interval of a function. They offer valuable insights into the behavior of the function on a local scale. To determine if a critical point is a local maximum or minimum, the second derivative test is used.

• Calculate the second derivative: \( y'' = 12x^2 - 4 \).
• Substitute critical points into the second derivative.

For our critical points:

• At \( x = 0 \), \( y'' = -4 \), indicating a local maximum (concave down).
• At \( x = 1 \) and \( x = -1 \), \( y'' = 8 \), indicating local minima (concave up).

This tells us how the function curves and the nature of the critical points.

Inflection Points

Inflection points occur where the function changes its concavity, which means the curve shifts from being concave up to concave down, or vice versa. At inflection points, the second derivative is zero, and there is a change in its sign. To find inflection points:

• Set the second derivative equal to zero: \( 12x^2 - 4 = 0 \).
• Solve to get \( x = \pm \sqrt{\frac{1}{3}} \).

Verify the change in concavity by selecting test points around \( x = \pm \sqrt{\frac{1}{3}} \):

• Check for sign changes of \( y'' \).

If a sign change occurs, these are inflection points, marking transitions in the curve's bending.

Derivative

The derivative of a function provides information about the slope of the tangent line at any given point. It shows how the function is changing. In this exercise, the derivative \( y' = 4x(x^2 - 1) \) helps us identify changes in the rate of the function.

• If \( y' > 0 \), the function is increasing.
• If \( y' < 0 \), the function is decreasing.

Finding the derivative is crucial for understanding the critical points and locating local extrema.

Second Derivative

The second derivative of a function shows the concavity of the function. It tells us how the rate of change of the slope is changing. Computes as \( y'' = 12x^2 - 4 \).
This derivative is a key part of identifying:

• Local maxima and minima using the second derivative test, where \( y'' > 0 \) indicates a local minimum and \( y'' < 0 \) indicates a local maximum.
• Inflection points, by identifying where the second derivative changes sign.

The second derivative gives us deeper insights into the curvature and acceleration of the function.