Problem 55 - Algebra and Trigonometry

Question

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 is 0.38 at x = -0.58; local minimum is -0.38 at x = 0.58.

Step 1: Find the first derivative

To find the local maximum and minimum, we need the derivative of the function. The first derivative of \(f(x) = x^3 - x\) is \(f'(x) = 3x^2 - 1\).

Step 2: Set the first derivative equal to zero

To find critical points, we set \(f'(x)\) equal to zero and solve for \(x\): \(3x^2 - 1 = 0\).

Step 3: Solve for x

Solve \(3x^2 - 1 = 0\):\[3x^2 = 1\]Divide by 3:\[x^2 = \frac{1}{3}\]Take the square root of both sides:\[x = \pm\frac{1}{\sqrt{3}} = \pm\frac{\sqrt{3}}{3}\]

Step 4: Find the second derivative

The second derivative of the function is used to determine the nature of the critical points. \(f''(x) = 6x\).

Step 5: Use the second derivative test

Evaluate \(f''(x)\) at the critical points to determine whether each is a local maximum or minimum.\[f''\left(\frac{\sqrt{3}}{3}\right) = 6 \cdot \frac{\sqrt{3}}{3} = 2\sqrt{3} > 0\]This indicates a local minimum.\[f''\left(-\frac{\sqrt{3}}{3}\right) = 6 \cdot -\frac{\sqrt{3}}{3} = -2\sqrt{3} < 0\]This indicates a local maximum.

Step 6: Compute function values at critical points

Calculate \(f(x)\) at the critical points to find the actual local maximum and minimum values. For \(x = \frac{\sqrt{3}}{3}\):\[f\left(\frac{\sqrt{3}}{3}\right) = \left(\frac{\sqrt{3}}{3}\right)^3 - \frac{\sqrt{3}}{3} = \frac{3\sqrt{3}}{27} - \frac{\sqrt{3}}{3} = -\frac{2\sqrt{3}}{9}\]For \(x = -\frac{\sqrt{3}}{3}\):\[f\left(-\frac{\sqrt{3}}{3}\right) = \left(-\frac{\sqrt{3}}{3}\right)^3 - \left(-\frac{\sqrt{3}}{3}\right) = -\frac{3\sqrt{3}}{27} + \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{9}\]

Step 7: State the local maxima and minima

The local maximum value is approximately \(0.38\) at \(x = -0.58\), and the local minimum value is approximately \(-0.38\) at \(x = 0.58\).

Key Concepts

Critical Points

In calculus, finding the critical points of a function is an essential step in determining where the function's graph is at its highest or lowest relative to nearby points. A critical point occurs where the first derivative of the function either equals zero or is undefined. This is because at these points, the slope of the tangent to the curve becomes flat, indicating potential peaks or valleys.

For the function \(f(x) = x^3 - x\), we start by calculating its first derivative \(f'(x) = 3x^2 - 1\). By setting the derivative equal to zero and solving, we can find the critical points:

• Set equation: \(3x^2 - 1 = 0\)
• Solving the above gives: \(x = \pm \frac{\sqrt{3}}{3}\)

These values represent the \(x\)-coordinates of the critical points, where potential local maxima or minima might occur. Understanding critical points helps us identify where the beginning of a curve change happens.

Derivatives

Derivatives are fundamental in calculus as they give you the rate at which a function is changing at any point. Think of the derivative as the function's version of a speedometer—telling us how the value of the function is increasing or decreasing.

When we derive the function \(f(x) = x^3 - x\), we obtain the first derivative \(f'(x) = 3x^2 - 1\). This equation helps us find the critical points by setting it equal to zero, revealing where the slope is zero and changes might occur.

Beyond finding critical points, understanding how derivatives work allows you to measure growth, decline, or steadiness in a wide range of real-world contexts:

• In physics, derivatives represent velocity and acceleration.
• In economics, derivatives can model profit changes relative to cost adjustments.
• Derivatives can also monitor things like temperature changes over time.

It’s the versatile tool for understanding change.

Second Derivative Test

The second derivative test is a strategy used to classify the nature of critical points found on a graph of a function based on its second derivative. By evaluating the second derivative at each critical point, we can determine whether each point is a maximum, minimum, or neither.

For our function \(f(x) = x^3 - x\), we first calculate the second derivative: \(f''(x) = 6x\). Next, we apply the second derivative test by evaluating this at each critical point:

• Critical point \(x = \frac{\sqrt{3}}{3}\): \(f''\left(\frac{\sqrt{3}}{3}\right) = 2\sqrt{3} > 0\), indicating a local minimum.
• Critical point \(x = -\frac{\sqrt{3}}{3}\): \(f''\left(-\frac{\sqrt{3}}{3}\right) = -2\sqrt{3} < 0\), indicating a local maximum.

Thus, the second derivative test is a critical tool in determining the concavity around critical points, helping us understand the shape and behavior of the function's graph.