Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(g(x)=\frac{x^{2}}{x^{3}+32}\) on \([0, \infty)\)

Step-by-Step Solution

Answer

The function decreases on \((0, 4)\) and increases on \((4, \infty)\) with a local minimum at \(x = 4\).

Step 1: Identify the Function

We are given the function \(g(x) = \frac{x^2}{x^3 + 32}\) to analyze on the interval \([0, \infty)\). This function is a rational function where the numerator is of degree 2 and the denominator is of degree 3.

Step 2: Domain of the Function

The function is defined for all values of \(x\geq0\), except where the denominator is zero. We solve the equation \(x^3 + 32 = 0\), which simplifies to \(x^3 = -32\). Since \(x\) is non-negative in the given interval and \(x^3\) cannot be negative, the denominator is never zero in this interval. Thus, the domain is \([0, \infty)\).

Step 3: Determine Asymptotes

For vertical asymptotes, we find values where the denominator is zero, which does not occur in \([0, \infty)\). For horizontal asymptotes, we look at the degree of the polynomial in the numerator and denominator. Since the degree of the denominator is higher (3 > 2), the horizontal asymptote is \(y = 0\).

Step 4: Evaluate Critical Points

Find the derivative using the quotient rule: \(g'(x) = \frac{(2x)(x^3 + 32) - (x^2)(3x^2)}{(x^3 + 32)^2}\). Simplifying gives \(g'(x) = \frac{2x^4 + 64x - 3x^4}{(x^3 + 32)^2} = \frac{-x^4 + 64x}{(x^3 + 32)^2}\). Set the numerator \(-x^4 + 64x = 0\) and solve for \(x\).

Step 5: Solve for Critical Points

Factor the equation \(-x^4 + 64x = 0\) as \(-x(x^3 - 64) = 0\). Solutions are \(x = 0\) and \(x = 4\). These are critical points within the interval \([0, \infty)\).

Step 6: Determine Intervals of Increase/Decrease

Using the critical points \(x = 0\) and \(x = 4\), test intervals \((0, 4)\) and \((4, \infty)\) in the derivative \(g'(x)\). For \(x = 2\) in \((0, 4)\), \(g'(x) < 0\), and for \(x = 5\) in \((4, \infty)\), \(g'(x) > 0\). So, the function is decreasing on \((0, 4)\) and increasing on \((4, \infty)\).

Step 7: Analyze Extrema

From the derivative test: at \(x = 4\), there's a minimum since \(g'(x)\) changes from negative to positive. At \(x = 0\), we test for endpoint and find \(g(0) = 0\). At \(x = 4\), \(g(4) = \frac{16}{48} = \frac{1}{3}\). Thus, \(x = 4\) is a local minimum.

Key Concepts

Critical Points

Critical points of a function like the one given, \(g(x) = \frac{x^2}{x^3 + 32}\), are essential for understanding its behavior. They are points where the function's derivative is zero or undefined. To find these points, we take the derivative of the function using the quotient rule. The derivative, found as \(g'(x) = \frac{-x^4 + 64x}{(x^3 + 32)^2}\), identifies where the rate of change is zero or possibly changes sign.
\[ -x^4 + 64x = 0 \]
Solving this gives us the critical points. These are the roots of the numerator: \(x = 0\) and \(x = 4\). By plugging these values into the derivative, we see where the function's slope changes from positive to negative and vice versa, marking potential local minima or maxima.

Asymptotes

Asymptotes help us understand the long-term behavior of rational functions like our given \(g(x)\). For this function, we focus on:

• Vertical asymptotes: Occur where the denominator equals zero. Since \(x^3 + 32 = 0\) has no real and positive solutions for \(x\) in the interval \([0, \infty)\), there are no vertical asymptotes in this domain.
• Horizontal asymptotes: Determined by looking at the degrees of the numerator and denominator. Here, the degree of the denominator (3) is higher than that of the numerator (2), leading us to a horizontal asymptote at \(y = 0\).

Horizontal asymptotes signify where the function stabilizes as \(x\) approaches infinity, providing a guide for sketching the function graph.

Intervals of Increase and Decrease

Determining where the function \(g(x)\) increases or decreases involves analyzing the sign of the derivative \(g'(x)\). Using critical points \(x = 0\) and \(x = 4\):

• Check the interval \((0, 4)\). If \(g'(x) < 0\), the function is decreasing.
• In the interval \((4, \infty)\), if \(g'(x) > 0\), the function is increasing.

Testing a number from each interval in the derivative, we found \(g(x)\) decreases on \((0, 4)\) and increases on \((4, \infty)\). This behavior indicates a local minimum at \(x = 4\), reflected in the overall graph ascent and descent.

Derivative Test

The derivative test is a method to determine local extrema, like minima or maxima, in a function. For \(g(x) = \frac{x^2}{x^3 + 32}\), the critical points found using the derivative help analyze such behavior. Initially, compute the derivative \(g'(x)\) with the quotient rule and set it equal to zero to find critical points.
\[ g'(x) = \frac{-x^4 + 64x}{(x^3 + 32)^2} \]
These critical points, \(x = 0\) and \(x = 4\), are plugged back into the intervals to test the sign changes in \(g'(x)\).If the derivative changes from negative to positive at a point, like \(x = 4\), it indicates a local minimum. Where \(g'(x)\) changes from positive to negative, there's a local maximum. However, in this problem, such a change only suggests a local minimum at \(x = 4\).