Problem 17 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. $$ g(x)=x-x^{3} $$

Step-by-Step Solution

Answer

The function \(g(x) = x - x^3\) is increasing on the interval \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\) and decreasing on the intervals \(x < -\frac{1}{\sqrt{3}}\) and \(x > \frac{1}{\sqrt{3}}\).

Step 1: Find the first derivative of the function g(x)

To find the first derivative, we will apply the power rule for differentiation. Given the function: \( g(x) = x - x^3 \) The first derivative is: \( g'(x) = 1 - 3x^2 \)

Step 2: Find the critical points

Critical points occur where the first derivative is zero or doesn't exist. In this case, g'(x) exists for all x, so we only need to find the zeros of g'(x). To do this, we set g'(x) equal to 0, and then solve for x: \( 1 - 3x^2 = 0 \\ 3x^2 = 1 \\ x^2 = \frac{1}{3} \\ x = \pm \frac{1}{\sqrt{3}} \) So the critical points are \(x=\frac{1}{\sqrt{3}}\) and \(x=-\frac{1}{\sqrt{3}}\).

Step 3: Test the intervals

To determine the intervals of increase and decrease, we will test the values of g'(x) in the intervals created by the critical points. We have three intervals to test: (1) \(x < -\frac{1}{\sqrt{3}}\) (2) \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\) (3) \(x > \frac{1}{\sqrt{3}}\) Choose a value in each interval and evaluate the first derivative: (1) \(x = -1, g'(-1) = 1 - 3(-1)^2 = 1 - 3 = -2 < 0\) (decreasing) (2) \(x = 0, g'(0) = 1 - 3(0)^2 = 1 > 0\) (increasing) (3) \(x = 1, g'(1) = 1 - 3(1)^2 = 1 - 3 = -2 < 0\) (decreasing)

Step 4: Write the intervals of increase and decrease

From the evaluations and analysis, we can now write the intervals where the function is increasing and decreasing: \(g(x) = x - x^3\) is increasing on the interval \( -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \), and decreasing on the intervals \(x < -\frac{1}{\sqrt{3}}\) and \(x > \frac{1}{\sqrt{3}}\).

Key Concepts

First Derivative Test

Understanding the first derivative test is foundational in calculus for determining where a function is increasing or decreasing. When we compute the first derivative of a function, we are essentially finding the slope at any point along that function. The sign of the slope—positive or negative—tells us whether the function is increasing or decreasing at that point.

Let's apply this to our exercise: We derived the first derivative of the function \( g(x) = x - x^3 \) and obtained \( g'(x) = 1 - 3x^2 \). The first derivative informs us about the rate of change of \( g(x) \). A positive rate of change (\( g'(x) > 0 \)) indicates the function is increasing, while a negative rate of change (\( g'(x) < 0 \)) indicates it is decreasing.

Using Test Points

After having identified the critical points, we divide the number line into intervals. We then select a test point from each interval to plug into the first derivative. For instance, choosing \( x = 0 \) for the second interval yields \( g'(0) = 1 > 0 \), showing that the function is increasing in that interval. In simpler terms, if you were hiking along the graph of \( g(x) \), between the critical points \(-\frac{1}{\sqrt{3}}\) and \(\frac{1}{\sqrt{3}}\), you'd be walking uphill.

Critical Points

A critical point is where the first derivative of a function is either zero or undefined, and these points are potential locations for local maxima or minima of the function. Finding critical points is a precursor to many calculus applications, such as sketching graphs and optimizing functions.

In our context, the critical points for \( g(x) = x - x^3 \) are solutions to \( g'(x) = 0 \). We found the critical points to be \(x = \pm \frac{1}{\sqrt{3}}\). These points split the number line into intervals where the function's behavior can change from increasing to decreasing, or vice versa.

Understanding Intervals

Imagine a critical point as a mountain peak or valley—the points where hikers have to decide whether to climb up or descend. In terms of our function, the intervals to the left and right of the peaks (in this case, the critical points) represent the different 'terrain' the graph goes through, whether that's an uphill or downhill slope.

Power Rule for Differentiation

The power rule is a quick and handy tool to differentiate functions where the variable has an exponent. In essence, the power rule states that if you have a function \( f(x) = x^n \), then its derivative is \( f'(x) = nx^{n-1} \). This allows us to differentiate polynomials term by term.

For our function \( g(x) = x - x^3 \), we used the power rule to differentiate each term separately. The derivative of \( x \) is \( 1 \) because the power rule dictates that \( nx^{n-1} \) becomes \( 1x^{1-1} = 1x^0 = 1 \). Similarly, for \( -x^3 \), we bring down the exponent as the coefficient and subtract one from the exponent to get \( -3x^2 \).

Applying the Power Rule

To apply the power rule efficiently, focus on treating each term independently. The power of a constant is zero, and any number raised to the power of zero is one. Hence, derivatives of constants are zero, which is why they disappear when using the power rule on polynomials like our function \( g(x) \). The power rule enables you to swiftly move through the differentiation process so you can reach the analysis of critical points and intervals of increase or decrease more quickly.