Problem 38 - Thomas Calculus

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 decreasing on \((- \infty, -2)\), increasing on \((-2, 0)\cup(0, \infty)\); local minimum at \(x = -2\), absolute minimum at \(x = 0\).

Step 1: Find the first derivative

To determine where the function \( g(x) = x^{2/3}(x+5) \) is increasing or decreasing, we first need to find its first derivative \( g'(x) \). Use the product rule for differentiation: \((u(x)v(x))' = u'(x)v(x) + u(x)v'(x)\), where \(u(x)=x^{2/3}\) and \(v(x)=x+5\). Calculate the derivatives: \(u'(x) = \frac{2}{3}x^{-1/3}\) and \(v'(x) = 1\). Thus, \(g'(x) = \frac{2}{3}x^{-1/3}(x+5) + x^{2/3}(1)\). Simplifying, we get \(g'(x) = \frac{2}{3}x^{-1/3}(x+5) + x^{2/3}\).

Step 2: Simplify the first derivative

To simplify \( g'(x) \), we write both terms with a common base: \(g'(x) = \frac{2(x+5)}{3x^{1/3}} + x^{2/3} = \frac{2x^{2/3} + 10x^{-1/3} + 3x^{2/3}}{3x^{1/3}}\). Combine terms: \(g'(x) = \frac{5x^{2/3} + 10x^{-1/3}}{3x^{1/3}}\). This can be further simplified to \(g'(x) = \frac{5x + 10}{3x^{1/3}}\).

Step 3: Find critical points

Critical points occur where \(g'(x) = 0\) or \(g'(x)\) is undefined. Set the numerator equal to zero: \(5x + 10 = 0\) gives \(x = -2\). \( g'(x) \) is also undefined at \(x = 0\) due to the \(x^{1/3}\) term in the denominator.

Step 4: Evaluate sign changes for intervals

Test intervals around the critical points \(x = -2\) and \(x = 0\). Check the sign of \(g'(x)\) in each interval: \((-\infty, -2)\), \((-2, 0)\), and \((0, \,\infty)\). For \(x < -2\), choose \(x = -3\): \(g'(-3) = \frac{5(-3) + 10}{3(-3)^{1/3}} < 0\), so \(g(x)\) is decreasing. For \(-2 < x < 0\), choose \(x = -1\): \(g'(-1) = \frac{5(-1) + 10}{3(-1)^{1/3}} > 0\), so \(g(x)\) is increasing. For \(x > 0\), choose \(x = 1\): \(g'(1) = \frac{5(1) + 10}{3(1)^{1/3}} > 0\), so \(g(x)\) is increasing.

Step 5: Determine intervals of increase and decrease

Based on the sign of \(g'(x)\) from Step 4, \(g(x)\) is decreasing on \((-\infty, -2)\) and increasing on \((-2, 0)\cup(0, \infty)\).

Step 6: Find critical function values

Evaluate \(g(x)\) at critical points: \(x = -2\) and \(x = 0\). Calculate \(g(-2) = (-2)^{2/3}\times(-2+5) = 3\) and \(g(0) = 0^{2/3}\times(0+5) = 0\).

Step 7: Identify local and absolute extrema

Since \(g(x)\) changes from decreasing to increasing at \(x = -2\), \(g(-2) = 3\) is a local minimum. At \(x = 0\), the function is increasing. The behavior at \(x \to -\infty\) and \(x \to \infty\) indicates the absence of other extrema. \(g(x)\) has a local minimum at \((x = -2, \ g(x) = 3)\) and an absolute minimum of \(0\) at \(x = 0\).

Key Concepts

Increasing and Decreasing Functions

Understanding where a function is increasing or decreasing is crucial in calculus, as it tells us about the behavior and trends of the graph. We determine this by examining the first derivative of the function. The first derivative, often denoted as \( f'(x) \) or \( g'(x) \) in this case, helps to determine the slopes of the tangent lines at any given point on the function. If \( g'(x) > 0 \), the function is increasing on that interval. This means the slope of the tangent line is positive, and the function's graph is moving upwards as you move from left to right. Conversely, if \( g'(x) < 0 \), the function is decreasing, meaning it is moving downwards as you travel from left to right.

For the function \( g(x) = x^{2/3}(x+5) \), the critical points found at \( x = -2 \) and \( x = 0 \) segment the interval into three sections: \((-abla\infty, -2)\), \((-2, 0)\), and \((0, \infty)\). To find out whether each of these intervals is increasing or decreasing:

• In \((-abla\infty, -2)\), the function is decreasing because \( g'(x) < 0 \).
• In \((-2, 0)\) and \((0, \infty)\), the function is increasing with \( g'(x) > 0 \).

This strategic analysis allows us to speculate on how the function will behave across its entire domain without needing to graph it entirely.

Critical Points

Critical points are where the first derivative of a function is zero or undefined. These points are important as they offer clues about potential extrema, which are points where the function could switch from increasing to decreasing, or vice versa.
A critical point occurs where \( g'(x) = 0 \) or is undefined, which can be seen in our example with \( g'(x) = \frac{5x + 10}{3x^{1/3}} \). To find these points, we solve:

• \( 5x + 10 = 0 \), resulting in \( x = -2 \).
• \( g'(x) \) is undefined at \( x = 0 \) due to the \( x^{1/3} \) in the denominator.

Once these critical points are identified, they become invaluable in determining the intervals where a function is increasing or decreasing and in finding any potential local extrema.

Local and Absolute Extrema

Local and absolute extrema are points where a function takes on a minimum or maximum value within a particular interval or on its entire domain.
For the function \( g(x) \), local extrema occur at points where \( g(x) \) changes direction, while an absolute extremum is the highest or lowest value over the entire domain.

• A local minimum is found at \( x = -2 \) with \( g(-2) = 3 \). This occurs because the function transitions from decreasing to increasing at this point.
• The absolute minimum occurs at \( x = 0 \) where \( g(0) = 0 \). In this case, the function has no local maximum, given that it smoothly increases beyond the critical points. As \( x \) approaches infinity, the function goes upwards without bound.

Understanding these extrema is essential in optimization problems and providing insights into the overall shape and peaks of the graph.