Problem 51 - Algebra and Trigonometry

Question

\(51-58=\) Find the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. $$ f(x)=x^{3}-x $$

Step-by-Step Solution

Answer

Local maximum at \(x=-0.58\) is 0.38; Local minimum at \(x=0.58\) is -0.38.

Step 1: Find the first derivative

To find the local maxima and minima, we first need to find the first derivative of the function. The given function is \( f(x) = x^3 - x \). The derivative, \( f'(x) \), is found using the power rule: \( f'(x) = 3x^2 - 1 \).

Step 2: Set the derivative equal to zero

To find the critical points, set the first derivative equal to zero and solve for \( x \). \[ 3x^2 - 1 = 0 \] Add 1 to both sides: \[ 3x^2 = 1 \] Divide both sides by 3: \[ x^2 = \frac{1}{3} \].

Step 3: Solve for x

Take the square root of both sides to solve for \( x \): \( x = \pm \sqrt{\frac{1}{3}} \). Therefore, the critical points are \( x = \frac{1}{\sqrt{3}} \) and \( x = -\frac{1}{\sqrt{3}} \).

Step 4: Find the second derivative

To determine whether each critical point is a local maximum or minimum, compute the second derivative \( f''(x) \) of \( f \). Therefore, the second derivative is \( f''(x) = 6x \).

Step 5: Evaluate the second derivative at critical points

Evaluate \( f''(x) \) at each critical point to apply the second derivative test. For \( x = \frac{1}{\sqrt{3}} \), \( f''\left(\frac{1}{\sqrt{3}}\right) = 6 \cdot \frac{1}{\sqrt{3}} \approx 3.46 \) which is positive, indicating a local minimum. For \( x = -\frac{1}{\sqrt{3}} \), \( f''\left(-\frac{1}{\sqrt{3}}\right) = 6 \cdot \left(-\frac{1}{\sqrt{3}}\right) \approx -3.46 \) which is negative, indicating a local maximum.

Step 6: Evaluate the function at the critical points

To find the local maximum and minimum values, substitute the critical points into the original function \( f(x) = x^3 - x \). For the local minimum at \( x = \frac{1}{\sqrt{3}} \), \( f\left(\frac{1}{\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}\right)^3 - \frac{1}{\sqrt{3}} \approx -0.38 \). For the local maximum at \( x = -\frac{1}{\sqrt{3}} \), \( f\left(-\frac{1}{\sqrt{3}}\right) = \left(-\frac{1}{\sqrt{3}}\right)^3 + \frac{1}{\sqrt{3}} \approx 0.38 \).

Key Concepts

Critical Points

Critical points are essential in understanding the behavior of a function. These are the values of \( x \) where the first derivative of the function equals zero or does not exist.
Critical points can indicate where a function changes from increasing to decreasing, meaning they could be locations of local maxima or minima. To find these points for the function \( f(x) = x^3 - x \), we start by finding the first derivative, \( f'(x) = 3x^2 - 1 \). Set \( f'(x) = 0 \) to find where the slope of the tangent to the curve is flat, indicating potential peaks or troughs.

• The equation comes out to \( 3x^2 - 1 = 0 \).
• Solve this to find \( x^2 = \frac{1}{3} \), which results in the critical points at \( x = \frac{1}{\sqrt{3}} \) and \( x = -\frac{1}{\sqrt{3}} \).

Identifying these points is the first step in further examining whether these are indeed points of local maximum or minimum.

First Derivative

The first derivative, \( f'(x) \), provides us information about the slope of the function. It tells us how the function is changing at any point \( x \). A positive derivative means the function is increasing at that point, while a negative derivative means it is decreasing.
Using the power rule, the derivative of our function \( f(x) = x^3 - x \) is determined to be \( f'(x) = 3x^2 - 1 \). To find potential local extrema, we look for where \( f'(x) = 0 \), meaning the curve has a horizontal tangent.

• The calculation of \( 3x^2 - 1 = 0 \) leads us to the critical points, \( x = \frac{1}{\sqrt{3}} \) and \( x = -\frac{1}{\sqrt{3}} \).

These points tell us where the function might have peaks or valleys, but additional tests are needed to confirm the nature of these points.

Second Derivative Test

The second derivative test uses the sign of the second derivative to determine the concavity of the function at the critical points. The concavity tells us whether the function opens up or down at these points.
For our function \( f(x) = x^3 - x \), the second derivative is \( f''(x) = 6x \). If \( f''(x) > 0 \), the function is concave up, and the critical point is a local minimum.
If \( f''(x) < 0 \), the function is concave down, and the critical point is a local maximum.

• At \( x = \frac{1}{\sqrt{3}} \), \( f''(\frac{1}{\sqrt{3}}) \approx 3.46 \), indicating a local minimum.
• At \( x = -\frac{1}{\sqrt{3}} \), \( f''(-\frac{1}{\sqrt{3}}) \approx -3.46 \), indicating a local maximum.

This test confirms whether the critical points reflect a peak or a dip in the function's graph.

Local Maximum and Minimum Values

The local maximum or minimum values of a function are the highest or lowest points in its immediate region. They are important because they help us understand the function's behavior in specific segments.
After identifying the critical points and confirming their nature using the second derivative test, we need to calculate the actual function values at these points to find the local extrema.
For the function \( f(x) = x^3 - x \):

• The local minimum value is \( f(\frac{1}{\sqrt{3}}) \approx -0.38 \), occurring at \( x = \frac{1}{\sqrt{3}} \).
• The local maximum value is \( f(-\frac{1}{\sqrt{3}}) \approx 0.38 \), occurring at \( x = -\frac{1}{\sqrt{3}} \).

These values provide a comprehensive view of the function around these critical points, finalizing the analysis of its local behavior.