Problem 30 - 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. $$ f(x)=x^{2 / 3}+5 $$

Step-by-Step Solution

Answer

The function \(f(x) = x^{\frac{2}{3}} + 5\) is decreasing on the interval \((-\infty, 0)\) and increasing on the interval \((0, +\infty)\).

Step 1: Find the first derivative of the function

To find out where the function is increasing or decreasing, we need to determine its first derivative. Using the power rule (where \(f(x) = x^n\), the derivative is \(f'(x) = nx^{n-1}\)) we have: $$ f'(x) = \frac{2}{3} x^{\frac{2}{3} - 1} $$ Now, simplify the exponent: $$ f'(x) = \frac{2}{3} x^{-\frac{1}{3}} $$

Step 2: Find the critical points

Critical points are where the first derivative is either equal to zero or undefined. In this case: $$ f'(x) = \frac{2}{3} x^{-\frac{1}{3}} $$ The denominator of the function is \(x^{\frac{-1}{3}}\), which means that the derivative will be undefined when \(x = 0\). Therefore, there is a critical point at \(x = 0\).

Step 3: Determine the intervals where the function is increasing and decreasing

We will now test the intervals around the critical point, \(x = 0\), to see if the function is increasing or decreasing. Choose a test point in each interval: 1. Interval: \((-\infty, 0)\) - Test point: \(x = -1\) Evaluate \(f'(-1)\): $$ f'(-1) = \frac{2}{3} (-1)^{-\frac{1}{3}} = -\frac{2}{3} < 0 $$ Since the derivative is negative in this interval, the function is decreasing on \((-\infty, 0)\). 2. Interval: \((0, +\infty)\) - Test point: \(x = 1\) Evaluate \(f'(1)\): $$ f'(1) = \frac{2}{3} (1)^{-\frac{1}{3}} = \frac{2}{3} > 0 $$ Since the derivative is positive in this interval, the function is increasing on \((0, +\infty)\). In conclusion, the function is decreasing on the interval \((-\infty, 0)\) and increasing on the interval \((0, +\infty)\).

Key Concepts

Derivatives

In calculus, derivatives help us understand how functions change. They show the rate at which a function is increasing or decreasing at any given point. The first derivative of a function, denoted as \( f'(x) \), gives us the slope of the tangent line to the curve at any point \( x \). A positive derivative indicates that the function is increasing, while a negative derivative indicates it is decreasing.
To find the derivative of \( f(x) = x^{2/3} + 5 \), we apply the power rule: for \( x^n \), the derivative is \( nx^{n-1} \). This results in:

• \( f'(x) = \frac{2}{3}x^{-1/3} \)

This simplified form of the derivative will help us analyze how the function behaves across different intervals of \( x \).

Critical Points

Critical points of a function occur where its first derivative is either zero or undefined. These points are significant because they can indicate potential maximums, minimums, or inflection points. For the function \( f(x) = x^{2/3} + 5 \), the first derivative is \( f'(x) = \frac{2}{3}x^{-1/3} \).
In this case, the derivative is undefined when \( x = 0 \). This occurs because \( x^{-1/3} \) introduces a division by zero at \( x = 0 \). Hence, \( x = 0 \) is a critical point.
Critical points require further testing to determine whether they correspond to maxima, minima, or points where the function simply changes direction (like increasing to decreasing, or vice-versa).

Increasing and Decreasing Functions

To identify where a function increases or decreases, we examine the sign of its first derivative. If the first derivative is positive in a particular interval, the function is increasing there. Conversely, if the derivative is negative, the function is decreasing. Let’s apply this to \( f(x) = x^{2/3} + 5 \):

• For the interval \((-\infty, 0)\), choose \( x = -1 \) as a test point. Evaluating \( f'(-1) \):
  \( f'(-1) = \frac{2}{3}(-1)^{-1/3} = -\frac{2}{3} \)
  This negative result shows the function decreases on this interval.
• For the interval \((0, +\infty)\), choose \( x = 1 \) as a test point. Evaluating \( f'(1) \):
  \( f'(1) = \frac{2}{3}(1)^{-1/3} = \frac{2}{3} \)
  Here, the positive result means the function increases on this interval.

Therefore, \( f(x) \) is decreasing on \((-\infty, 0)\) and increasing on \((0, +\infty)\). This analysis allows us to understand the function’s behavior across its domain.