Problem 43 - Calculus with Concepts in Calculus

Question

Use the First Derivative Test or the Second Derivative Test to determine the relative extreme values, if any, of the function. Then sketch the graph of the function. $$ f(x)=\left(x^{2}-1\right)^{2} $$

Step-by-Step Solution

Answer

The function has a maximum at \( x = 0 \) and minima at \( x = -1 \) and \( x = 1 \).

Step 1: Find the derivative of the function

To apply the First or Second Derivative Test, we first need the derivative of the function. The function is \( f(x) = (x^2 - 1)^2 \). Using the chain rule, the derivative is \( f'(x) = 2(x^2 - 1) \cdot 2x = 4x(x^2 - 1) \).

Step 2: Find critical points

Critical points occur where the first derivative \( f'(x) \) is zero or undefined. In this case, we set \( 4x(x^2 - 1) = 0 \). Solving this gives \( x = 0 \), \( x = 1 \), and \( x = -1 \) as critical points.

Step 3: Apply the First Derivative Test

To determine whether these critical points are relative extrema, consider the sign of \( f'(x) \) around each point. For \( x < -1 \), \( f'(x) < 0 \). For \( -1 < x < 0 \), \( f'(x) > 0 \). For \( 0 < x < 1 \), \( f'(x) < 0 \). For \( x > 1 \), \( f'(x) > 0 \). These sign changes indicate that \( x = -1 \) is a relative minimum, \( x = 0 \) is a relative maximum, and \( x = 1 \) is a relative minimum.

Step 4: Second Derivative for Confirmation

Alternatively, we can use the second derivative for confirmation. The second derivative is \( f''(x) = 4(3x^2 - 1) \). Evaluating at the critical points: \( f''(-1) = 8 \), which is positive, confirming a minimum at \( x = -1 \). \( f''(0) = -4 \), which is negative, confirming a maximum at \( x = 0 \). \( f''(1) = 8 \), which is positive, confirming a minimum at \( x = 1 \).

Step 5: Sketch the graph

Based on the analysis, the graph of the function has minima at \( x = -1 \) and \( x = 1 \), and a maximum at \( x = 0 \). At these points, the function values can be calculated: \( f(-1) = 0 \), \( f(0) = 1 \), \( f(1) = 0 \). The graph is symmetric about the y-axis due to the even power of the function.

Key Concepts

Second Derivative Test

The Second Derivative Test is a fundamental tool in calculus to determine the concavity of a function at its critical points. It offers a straightforward way to confirm whether those points are relative minima or maxima. To perform this test, we first need the function's second derivative. Let's review: the original function is given as \( f(x) = (x^2 - 1)^2 \). When the second derivative, \( f''(x) \), is evaluated at the critical points, we observe:

• If \( f''(c) > 0 \), the point \( x = c \) is a local minimum since the function is concave up at that point.
• If \( f''(c) < 0 \), the point \( x = c \) is a local maximum because the function is concave down.
• If \( f''(c) = 0 \), the test is inconclusive, and we might need other methods for confirmation, such as re-evaluating the first derivative test.

In this exercise, \( f''(-1) = 8 \) and \( f''(1) = 8 \) both being positive indicate minima at \( x = -1 \) and \( x = 1 \). Meanwhile, \( f''(0) = -4 \) shows a negative value, confirming a maximum at \( x = 0 \). Thus, the second derivative test acts as an effective cross-check after we've identified critical points.

critical points

Critical points in a function are essential to understanding where potential extrema can occur. These are the values of \( x \) where the derivative of the function is zero or undefined. For the given function \( f(x) = (x^2 - 1)^2 \), its first derivative is \( f'(x) = 4x(x^2 - 1) \). To find critical points:

• Set the derivative equal to zero: \( 4x(x^2 - 1) = 0 \).
• Solve for \( x \), yielding \( x = 0 \), \( x = 1 \), and \( x = -1 \).

Critical points serve as the potential locations for relative maxima or minima. It's essential to apply further tests, like the first or second derivative tests, to determine the nature of these points as extrema.

relative extrema

Relative extrema refer to the peaks and valleys of a function on a given interval where the function changes direction, forming either a maximum or minimum point. Once critical points are identified, analyzing the sign of the derivative around these points helps in classifying them as relative minima or maxima. For the exercise function, we classified its critical points as:

• Relative minimum at \( x = -1 \) and \( x = 1 \), observed where the function dips lowest before increasing.
• Relative maximum at \( x = 0 \), seen as the top of a peak where the function increases and then decreases.

Graphically, these extrema shape the broad features of the function's graph. Understanding where these occur allows students to sketch accurate graphs and gain insights into the function's behavior.

derivative of a function

The derivative of a function is a core concept in calculus, describing the rate at which the function's value changes as its input changes. For the exercise function, the derivative \( f'(x) = 4x(x^2 - 1) \) tells us how the function \( f(x) = (x^2 - 1)^2 \) behaves at every point along the x-axis.

• This derivative is calculated using the chain rule, which is essential when dealing with composite functions.
• The first derivative provides crucial information on where the function is increasing or decreasing based on its sign.

Understanding derivatives enables students to find intervals of growth or decline for functions, locate turning points, and frame the overall slope of the curve. This leads to deeper insights into the nature of the curve and how it will look visually through sketching.