Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(g(x)=x^{2}+\frac{16 x^{2}}{(8-x)^{2}}\) on \((8, \infty)\)

Step-by-Step Solution

Answer

The function \( g(x) \) increases on \( (8, \infty) \) and diverges to infinity as \( x \) approaches 8 or \( \infty \).

Step 1: Simplify the function

The given function is \( g(x) = x^{2} + \frac{16x^{2}}{(8-x)^{2}} \). The first term is already simplified, so let's focus on the second term: \( \frac{16x^{2}}{(8-x)^{2}} \). There is no further simplification possible since it's already in its simplest form. The function remains \( g(x) = x^{2} + \frac{16x^{2}}{(8-x)^{2}} \).

Step 2: Identify the domain

The function is defined for \( x > 8 \) since dividing by zero is not allowed and \( x = 8 \) would make the denominator \( (8-x)^{2} = 0 \). The domain of \( g(x) \) is therefore \((8, \infty)\).

Step 3: Find critical points

To find critical points, take the derivative of \( g(x) \) and set it to zero. First, differentiate each term separately:- The derivative of \( x^2 \) is \( 2x \).- Apply the quotient rule to differentiate \( \frac{16x^2}{(8-x)^2} \): \[ \frac{d}{dx}\left(\frac{16x^2}{(8-x)^2}\right) = \frac{(32x(8-x)^2 + 16x^2\times2(8-x))}{(8-x)^4} \]Combine terms and simplify to find where the derivative is zero.

Step 4: Solve derivative equation

Solve the equation from the derivative computed: - Simplify and solve for \( x \) in the equation: \[ 32x(8-x)^2 - 32x^2(8-x) = 0 \]- Factor out the common terms and solve for \( x \).This will yield the critical points for \( x > 8 \).

Step 5: Test second derivative

Calculate the second derivative to test whether the critical points are minima, maxima, or points of inflection.If the second derivative \( g''(x) \) evaluated at the critical points is positive, the function has a local minimum. If negative, it has a local maximum. If zero, further test is needed.

Step 6: Analyze behavior at boundaries

Consider the behavior of the function as \( x \to 8^+ \) and \( x \to \infty \).- As \( x \to 8^+ \), \( (8-x)^2 \to 0 \) causes the second term \( \frac{16x^2}{(8-x)^2} \) to approach infinity, driving \( g(x) \to \infty \).- As \( x \to \infty \), the \( \frac{16x^2}{(8-x)^2} \) term approaches 0, so \( g(x) \) behaves like \( x^2 \).

Key Concepts

Derivative

A derivative in calculus represents how a function changes as its input changes. It's the slope of the function at any given point. For the function \( g(x) = x^2 + \frac{16x^2}{(8-x)^2} \), finding the derivative involves differentiating each part separately. The derivative of \( x^2 \) is \( 2x \).

The second term, \( \frac{16x^2}{(8-x)^2} \), requires us to use the quotient rule, which is used when differentiating functions of the form \( \frac{f(x)}{g(x)} \). The quotient rule states: \[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]

By applying this rule, we differentiate the top (\( f(x) = 16x^2 \)) and bottom (\( g(x) = (8-x)^2 \)) separately. This process finds the rate at which \( g(x) \) changes and tells us where its critical points might be.

Critical Points

Critical points provide valuable information about a function's behavior. These are the points where the function's derivative is zero or undefined, indicating potential maximum, minimum, or inflection points.

To find them, set the derivative \( g'(x) \) of the function equal to zero and solve for \( x \). This process helps determine where the function stops increasing or decreasing, which is crucial for understanding the shape of the graph.

Once you have the derivative set up, simplify and solve the resulting equation to find these critical points. Remember, only consider points where \( x > 8 \) due to the function's domain constraints.

Second Derivative

The second derivative provides insight into the concavity and shape of the function's graph. It is the derivative of the derivative. This means it measures how the rate of change itself is changing.

For instance, if the second derivative \( g''(x) \) is positive at a point, the function is concave up at that point, indicating a local minimum. Conversely, if \( g''(x) \) is negative, the function is concave down, suggesting a local maximum. Checking the second derivative at critical points helps confirm whether these points are indeed local maxima or minima.

In summary, the second derivative helps us look deeper into the function’s geometry, revealing subtler details not apparent from the first derivative alone.

Domain

The domain of a function is the set of all possible input values (\( x \)) that allow the function to work without any undefined behaviors like division by zero. In our function, \( g(x) = x^2 + \frac{16x^2}{(8-x)^2} \), the term \( \frac{16x^2}{(8-x)^2} \) becomes problematic at \( x=8 \), since this would make the denominator zero.

Thus, the domain is \( (8, \infty) \), meaning \( x \) can be any value greater than 8. Understanding the domain is crucial as it informs us about the limits within which we can validly analyze the function.