Problem 35 - Algebra and Trigonometry

Question

\(35-42\) A function is given. (a) Find all 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. (b) Find the intervals on which the function is increasing and on which the function is decreasing. State each answer correct to two decimal places. $$ f(x)=x^{3}-x $$

Step-by-Step Solution

Answer

Local max at \( x = -0.577 \), \( f(x) = 0.385 \); local min at \( x = 0.577 \), \( f(x) = -0.385 \). Increasing on \((-\infty, -0.577) \cup (0.577, \infty)\); decreasing on \((-0.577, 0.577)\).

Step 1: Find the First Derivative

To determine the critical points, we first need to find the derivative of the function. The function is given as \( f(x) = x^3 - x \). Taking the derivative, we get: \[ f'(x) = 3x^2 - 1. \]

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find critical points: \[ 3x^2 - 1 = 0. \]Rearranging, we have:\[ 3x^2 = 1 \]\[ x^2 = \frac{1}{3} \]\[ x = \pm \sqrt{\frac{1}{3}}. \]Solving, we get:\[ x = \pm 0.577. \]

Step 3: Find the Second Derivative

The second derivative test helps determine whether each critical point is a local maximum, local minimum, or neither. Find the second derivative:\[ f''(x) = 6x. \]

Step 4: Apply Second Derivative Test

Evaluate the second derivative at each critical point: - For \( x = 0.577 \): \[ f''(0.577) = 6(0.577) = 3.462 \] (positive, indicating a local minimum).- For \( x = -0.577 \): \[ f''(-0.577) = 6(-0.577) = -3.462 \] (negative, indicating a local maximum).

Step 5: Calculate Function Values at Critical Points

Determine the values of the function at the critical points:- For \( x = 0.577 \): \[ f(0.577) = (0.577)^3 - 0.577 = -0.385. \]- For \( x = -0.577 \): \[ f(-0.577) = (-0.577)^3 - (-0.577) = 0.385. \]

Step 6: Determine Intervals of Increase and Decrease

Use the sign of the first derivative to find intervals of increase and decrease:- For \( f'(x) = 3x^2 - 1 \), solve the inequality: - When \( f'(x) > 0 \), the intervals are increasing: \( x < -0.577 \) or \( x > 0.577 \). - When \( f'(x) < 0 \), the interval is decreasing: \( -0.577 < x < 0.577 \).

Key Concepts

Local Maximum and Minimum

In calculus, finding the local maximum and minimum of a function involves identifying where the function reaches its highest or lowest values, respectively, within a particular interval.
These points are crucial in understanding the behavior of the function in a specific region. For our function, given as \( f(x) = x^3 - x \), we employ derivatives to spot these notes.

The second derivative test is often used after determining the critical points where the first derivative is zero. Specifically, when the second derivative is positive at a point, that point is a local minimum. Conversely, if the second derivative is negative at a point, that point qualifies as a local maximum.

Here, \( x = 0.577 \) results in a positive second derivative, thus is a local minimum, and \( x = -0.577 \) provides a negative second derivative, serving as a local maximum. Understanding these points allows us to sketch the function's graph beautifully with correct peaks and troughs.

First Derivative Test

The first derivative test is a valuable method in calculus to analyze the behavior of a function and determine where it has turning points, which could be potential local maxima or minima.
It primarily involves assessing changes in sign of the first derivative \( f'(x) \), which is derived by differentiating the original function.

Consider our function: \( f(x) = x^3 - x \). Its first derivative, \( f'(x) = 3x^2 - 1 \), serves as the key tool for this test.

We set \( f'(x) \) to zero to find the critical points \( x = \pm 0.577 \). Afterwards, examining the intervals around these points indicates whether the function changes from increasing to decreasing or vice versa.

• If \( f'(x) \) shifts from positive to negative, it's a local maximum.
• If \( f'(x) \) shifts from negative to positive, it's a local minimum.

This process gives insight into the increasing or decreasing nature of the function's curve.

Second Derivative Test

The second derivative test is a quick and effective way to determine whether a function's critical points are local maxima, minima, or points of inflection.
It involves calculating the second derivative, denoted as \( f''(x) \), which tells us about the concavity of the function.

In this exercise, the function \( f(x) = x^3 - x \) has a second derivative of \( f''(x) = 6x \). Evaluating it at the critical points \( x = 0.577 \) and \( x = -0.577 \) allows us to classify these points easily:

• \( f''(0.577) > 0 \) indicates a local minimum because the curve is concave upward.
• \( f''(-0.577) < 0 \) signifies a local maximum given that the curve is concave downward.

Using this test simplifies the classification of points without extra calculations or graph plotting, making it an efficient decision-making tool in calculus.

Critical Points

Critical points are special places on a graph where the derivative of the function is zero or undefined, typically where the function tips over before increasing or decreasing again.
Discovering them is pivotal for analyzing a function's graph to pinpoint its peaks, troughs, and inflection points.

Let's apply this to the function \( f(x) = x^3 - x \). We find its critical points by setting the first derivative \(f'(x) = 3x^2 - 1\) equal to zero. Solving this, we get \( x = \pm 0.577 \).

These critical points demonstrate where changes in the function's growth rate occur. They are useful for understanding sections of the graph where the function might have valleys (local minima) or peaks (local maxima), which are essential for creating a full picture of the curve's geometry.