Problem 37 - 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. $$f(x)=x^{1 / 3}(x+8)$$

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, -2)\) and \((-2, \infty)\). No local or absolute extrema.

Step 1: Understand the Problem

We need to determine the intervals where the function \( f(x) = x^{1/3}(x+8) \) is increasing or decreasing, and also find any local or absolute extremums (minimums or maximums). This involves finding the first derivative and analyzing its critical points.

Step 2: Find the Derivative

To find where the function is increasing or decreasing, we first calculate its derivative. Use the product rule for differentiation, given two functions \( u(x) = x^{1/3} \) and \( v(x) = x+8 \), the derivative is: \[ f'(x) = u'(x)v(x) + u(x)v'(x) = \frac{1}{3}x^{-2/3}(x+8) + x^{1/3} \cdot 1 \] After simplifying, \[ f'(x) = \frac{x+8}{3x^{2/3}} + x^{1/3} \].

Step 3: Simplify and Find Critical Points

Combine terms under a common denominator:\[ f'(x) = \frac{x+8 + 3x}{3x^{2/3}} = \frac{4x+8}{3x^{2/3}} \].Set \( f'(x) = 0 \) to find critical points: \[ 4x + 8 = 0 \Rightarrow x = -2 \].

Step 4: Determine Intervals of Increase and Decrease

To determine where the function increases or decreases, evaluate \( f'(x) \) around the critical point \( x = -2 \):- For \( x < -2 \), choose \( x = -3 \), then \( f'(-3) = \frac{-12 + 8}{-9} > 0 \), the function is increasing.- For \( x > -2 \), choose \( x = 0 \), then \( f'(0) = \frac{8}{0} > 0 \), the function is still increasing.So, increase interval is \( (-\infty, -2) \cup (-2, \infty) \) and no decrease.

Step 5: Identify Local and Absolute Extremes

Since \( f(x) \) continually increases in serviceable domains except the point of discontinuity at \( x = 0 \), there are no local maxima or minima. No absolute extreme exists in the increased domain due to continuous progression incrementally.

Key Concepts

Derivative

In calculus, the derivative of a function provides crucial information about the function's behavior, such as its rate of change. The derivative is essentially a formula that tells us how the output of a function changes as its input varies. Calculating the derivative involves rules like the product rule, chain rule, and quotient rule, depending on the structure of the function.

For the given function, the product rule was used to find the derivative. If you have two functions multiplied together, say \( u(x) \) and \( v(x) \), the product rule states:

• \( (uv)' = u'v + uv' \)

This rule was applied to differentiate \( f(x) = x^{1/3}(x+8) \). The derivative \( f'(x) = \frac{x+8}{3x^{2/3}} + x^{1/3} \) is key for further analysis, such as determining where the function increases or decreases.

Increasing and Decreasing Intervals

Increasing and decreasing intervals describe where a function rises or falls as you move along the x-axis. To find these intervals, we need to examine the sign of the derivative. When \( f'(x) > 0 \), the function is increasing; when \( f'(x) < 0 \), it is decreasing.

For example, once we find the derivative \( f'(x) = \frac{4x+8}{3x^{2/3}} \), we can test values in different intervals to see whether the derivative is positive or negative.

• For \( x < -2 \), the derivative is positive, indicating that the function is increasing.
• For \( x > -2 \), the function remains increasing because the derivative stays positive.

This tells us that there's no interval where the function is genuinely decreasing, leading to a consistent rising pattern except at critical points.

Critical Points

Critical points are where the derivative is zero or undefined, often indicating potential minima or maxima. They are special points on the graph of a function where the tangent is either horizontal or vertical. To find critical points, set the derivative equal to zero.

For the function \( f(x) = x^{1/3}(x+8) \), the critical point was found by solving \( f'(x) = \frac{4x+8}{3x^{2/3}} = 0 \), leading to \( x = -2 \). Here, the change from one behavior (increasing) to another is tested, even though in this case, the function continues to increase past this point. There is also a point of discontinuity at \( x = 0 \), though not critical in determining increases or decreases in the function, it represents a vertical tangent, indicating where normal behavior analysis stops.

Extreme Values

Extreme values refer to the highest or lowest points on a function within a given interval. These can be local (local maxima/minima) or global (absolute maximum/minimum) extremes. Extreme values occur at critical points or endpoints of a domain where the function can achieve high or low output values.

In this particular case, the function continually increases on either side of the critical point \( x = -2 \), with no switching to decreasing, suggesting there are no local maxima or minima. Since the domain continues to infinity, no absolute extreme values are present as well. Understanding extreme values help in comprehending the broad structure of a function, such as potential constraints in optimization problems.