Problem 38 - Thomas Calculus in SI Units

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$g(x)=x^{2 / 3}(x+5)$$

Step-by-Step Solution

Answer

g(x) is increasing on \(( -\infty, -2) \cup (0, \infty)\) and decreasing on \((-2, 0)\). Local max at \(x = -2\), min at \(x = 0\).

Step 1: Find the First Derivative

To determine the intervals of increase and decrease, we first find the derivative of the function, \( g(x) = x^{2/3}(x+5) \). Using the product rule, we have:\[g'(x) = \frac{2}{3}x^{-1/3}(x+5) + x^{2/3}\]This simplifies to:\[g'(x) = \frac{2}{3}x^{-1/3}(x+5) + x^{2/3}\]We need to simplify it to find critical points.

Step 2: Simplify the Derivative

Simplify \( g'(x) \):\[g'(x) = \frac{2}{3} \cdot x^{-1/3} \cdot (x+5) + x^{2/3}\]Factor out \( x^{-1/3} \):\[g'(x) = x^{-1/3} \left( \frac{2}{3}(x+5) + x^{1} \right)= x^{-1/3} \left( \frac{2}{3}x + \frac{10}{3} + x\right)= x^{-1/3} \left(\frac{5x}{3} + \frac{10}{3}\right)\]

Step 3: Find Critical Points

Set \( g'(x) = 0 \) to find critical points:\[x^{-1/3} \left(\frac{5x}{3} + \frac{10}{3}\right) = 0\]Solving for \( x \), we have:\( x = 0 \) or \( \frac{5x}{3} + \frac{10}{3} = 0 \), leading to:\[5x + 10 = 0 \Rightarrow x = -2\]So, the critical points are \( x = 0 \) and \( x = -2 \).

Step 4: Determine Intervals Using Test Points

We evaluate \( g'(x) \) using test points in the intervals determined by the critical points \( x = -2 \) and \( x = 0 \):- For \( x < -2 \), choose \( x = -3 \): \( g'(-3) > 0 \), so \( g(x) \) is increasing.- For \( -2 < x < 0 \), choose \( x = -1 \): \( g'(-1) < 0 \), so \( g(x) \) is decreasing.- For \( x > 0 \), choose \( x = 1 \): \( g'(1) > 0 \), so \( g(x) \) is increasing.

Step 5: Identify Local and Absolute Extrema

Evaluate \( g(x) \) at critical points to determine local extrema:\[g(-2) = (-2)^{2/3}((-2)+5) = 3g(0) = 0^{2/3}(0+5) = 0\]Since \( g(x) \) changes from increasing to decreasing at \( x = -2 \), there is a local maximum at \( x = -2 \). At \( x = 0 \), \( g(x) \) changes from decreasing to increasing, so there is a local minimum at \( x = 0 \).

Key Concepts

Calculus

Calculus is a branch of mathematics that focuses on change. It primarily deals with derivatives and integrals. Derivatives help us understand how a function changes at any given point. Integrals, on the other hand, are used to find areas under curves.
Functions like \(g(x) = x^{2/3}(x+5)\) are analyzed in calculus to determine their behavior. Calculus helps to identify where such functions increase or decrease, and where they have peaks and valleys. These insights are valuable in physics, engineering, economics, and many other fields.

Derivative Test

The derivative test is crucial in calculus for finding critical points and analyzing function behavior. The critical points of a function occur where its derivative is zero or undefined. For our function, the first derivative, \(g'(x)\), helps identify these points:

• First, calculate the derivative using rules like the product rule.
• Then, simplify it to find expressions that help locate critical points.

In this situation, critical points are where \(g'(x) = 0\). These points, like \(x = 0\) and \(x = -2\) for our function, inform us where the direction of the graph might change from increasing to decreasing or vice versa.

Intervals of Increase and Decrease

Intervals of increase and decrease show where a function goes up or down, respectively. To identify these intervals, look at the sign of the derivative:

• If \(g'(x) > 0\), the function is increasing in that interval.
• If \(g'(x) < 0\), the function is decreasing.

For \(g(x) = x^{2/3}(x+5)\), we use these rules:

• Before \(x = -2\), the function is increasing.
• Between \(x = -2\) and \(x = 0\), it is decreasing.
• After \(x = 0\), it increases again.

Analyzing intervals is essential for understanding the overall shape and behavior of graphs.

Local and Absolute Extrema

Local and absolute extrema refer to the highest or lowest points a function reaches locally (near a point) and overall (in its domain).
Local extrema occur at critical points where the derivative changes sign:

• For \(g(x) = x^{2/3}(x+5)\), local maxima occur at \(x = -2\) because the function changes from increasing to decreasing.
• Local minima occurs at \(x = 0\) as it changes back to increasing.

Absolute extrema require evaluating the function over its entire domain, considering endpoints and infinity if applicable. For this function, without endpoints or bounds, the critical points provide the necessary information for local behavior.