Problem 38 - 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. $$ f(x)=x^{3}-3 x $$

Step-by-Step Solution

Answer

The function has a relative maximum at \((-1, 2)\) and a relative minimum at \((1, -2)\).

Step 1: Find the First Derivative

To determine the critical points of the function, we first need to find its derivative. Start by differentiating the function \( f(x) = x^3 - 3x \). The derivative is given by: \( f'(x) = 3x^2 - 3 \).

Step 2: Set the First Derivative to Zero

To find critical points, set \( f'(x) \) to zero and solve for \( x \).\[ 3x^2 - 3 = 0 \]\[ 3(x^2 - 1) = 0 \] \[ x^2 - 1 = 0 \] This factors to \( (x-1)(x+1) = 0 \), giving us the critical points \( x = 1 \) and \( x = -1 \).

Step 3: Apply the First Derivative Test

To determine the nature of the critical points, apply the First Derivative Test. Compute \( f'(x) \) around the critical points, using test points, such as for intervals \((-\infty, -1)\), \((-1, 1)\), and \((1, \infty)\). - For \( x < -1 \), pick \( x = -2 \): \( f'(-2) = 3(-2)^2 - 3 = 9 > 0 \).- For \( -1 < x < 1 \), pick \( x = 0 \): \( f'(0) = 3(0)^2 - 3 = -3 < 0 \).- For \( x > 1 \), pick \( x = 2 \): \( f'(2) = 3(2)^2 - 3 = 9 > 0 \).Since \( f'(x) \) changes from positive to negative at \( x = -1 \), there is a relative maximum at \( x = -1 \). It changes from negative to positive at \( x = 1 \), indicating a relative minimum.

Step 4: Calculate Function Values at Critical Points

To find the actual relative extrema, calculate \( f(x) \) at these critical points:- \( f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2 \).- \( f(1) = (1)^3 - 3(1) = 1 - 3 = -2 \). Thus, the relative maximum is at \((-1, 2)\) and the relative minimum is at \((1, -2)\).

Step 5: Sketch the Graph

Now use this information to sketch the graph of \( f(x) = x^3 - 3x \). The graph passes through the origin, has a relative maximum at \((-1, 2)\), and a relative minimum at \((1, -2)\). It increases on \((-\infty, -1)\), decreases on \((-1, 1)\), and increases again on \((1, \infty)\).

Key Concepts

First Derivative Test

The First Derivative Test is a useful tool for determining the relative extrema of a function, which are points where the function changes from increasing to decreasing or vice versa. Here’s how it works:

First, find the critical points of the function by setting the first derivative to zero. For the function \(f(x) = x^3 - 3x\), its derivative is \(f'(x) = 3x^2 - 3\). Setting \(f'(x)\) to zero gives us the equation \(3(x^2 - 1) = 0\), which results in critical points at \(x = 1\) and \(x = -1\).

Once you have the critical points, apply the First Derivative Test by checking the sign of the derivative on intervals around these points. Pick test points such as \(x = -2, 0,\) and \(2\):

• \(f'(-2) = 9 > 0\), so \(f(x)\) is increasing on \((-\infty, -1)\).
• \(f'(0) = -3 < 0\), so \(f(x)\) is decreasing on \((-1, 1)\).
• \(f'(2) = 9 > 0\), so \(f(x)\) is increasing on \((1, \infty)\).

Hence, the function changes from increasing to decreasing at \(x = -1\), indicating a relative maximum, and from decreasing to increasing at \(x = 1\), indicating a relative minimum.

Second Derivative Test

The Second Derivative Test can also help identify the nature of critical points without analyzing the sign changes of the first derivative. It relies on the sign of the second derivative, \(f''(x)\), at the critical points.

To use the Second Derivative Test, follow these steps:

• Compute the second derivative of the function. For \(f(x) = x^3 - 3x\), the second derivative is \(f''(x) = 6x\).
• Evaluate \(f''(x)\) at each critical point.

For \(x = -1\):
\(f''(-1) = 6(-1) = -6 < 0\), indicating a relative maximum because the curvature of the graph is concave down at this point.

For \(x = 1\):
\(f''(1) = 6(1) = 6 > 0\), indicating a relative minimum because the curvature is concave up at this point.

This test provides a quick way to determine the type of extrema without plotting intervals.

Critical Points

Critical points are where the derivative of a function is zero or undefined. These points are crucial when trying to identify potential maxima, minima, or points of inflection on the graph of a function. For the cubic function \(f(x) = x^3 - 3x\), we found the critical points by first computing the derivative \(f'(x) = 3x^2 - 3\) and setting it to zero:

\[3(x^2 - 1) = 0\]
This factors to \((x-1)(x+1) = 0\), yielding the critical points \(x = 1\) and \(x = -1\).

It's important to check these critical points against the First and Second Derivative Tests to determine their nature – whether they're locations of relative maxima, minima, or neither. Identifying critical points is the first essential step in analyzing any function to understand its behavior and graph.

Relative Extrema

Relative extrema refer to the relative maximum and minimum points on a function. These are the highest and lowest points in a particular interval, though not necessarily the highest or lowest overall.

For the function \(f(x) = x^3 - 3x\), after using the First and Second Derivative Tests, the relative extrema were found:

• The relative maximum occurs at \(x = -1\), with \(f(-1) = 2\).
• The relative minimum occurs at \(x = 1\), with \(f(1) = -2\).

These points indicate where the direction of the graph changes—an increase turns to decrease at a maximum, and a decrease turns to increase at a minimum. When graphing the function, these points help define the shape and behavior of the curve, making them essential for a complete understanding of the function's characteristics.