Problem 53 - Calculus Single Variable

Question

In each of Exercises 48-53, a function \(f\) is given. In each case, the domain of \(f\) and the value of \(f(x)\) depend on a positive constant \(b\). Determine each point at which \(f\) has a local extremum. Use the First Derivative Test to classify the extremum. $$ f(x)=\left(x^{4}+8 b^{4}\right) /\left(x^{2}+b^{2}\right), b

Step-by-Step Solution

Answer

The function has no local extremum within the specified domain due to boundary overlaps at possible critical point evaluation to \( x = b \).

Step 1: Understand the function and its domain

The function given is \( f(x) = \frac{x^4 + 8b^4}{x^2 + b^2} \). The domain of \( f \) is determined by the condition \( b < x \), which implies that \( x \) must be greater than the positive constant \( b \). Hence, the domain is \( x > b \).

Step 2: Find the derivative \( f'(x) \)

To find the derivative, use the quotient rule: \( f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2} \) where \( g(x) = x^4 + 8b^4 \) and \( h(x) = x^2 + b^2 \). Therefore, \( g'(x) = 4x^3 \) and \( h'(x) = 2x \). Hence, the derivative is:\[ f'(x) = \frac{(4x^3)(x^2 + b^2) - (x^4 + 8b^4)(2x)}{(x^2 + b^2)^2} \].

Step 3: Simplify the derivative

Simplify the expression for \( f'(x) \):- Distribute and combine terms in the numerator:\[ 4x^3(x^2 + b^2) = 4x^5 + 4x^3b^2 \]\[ 2x(x^4 + 8b^4) = 2x^5 + 16xb^4 \]- Combine them:\[ f'(x) = \frac{4x^5 + 4x^3b^2 - 2x^5 - 16xb^4}{(x^2 + b^2)^2} \]- Further simplify:\[ f'(x) = \frac{2x^5 + 4x^3b^2 - 16xb^4}{(x^2 + b^2)^2} \].

Step 4: Find critical points

Set the simplified derivative equal to zero and solve for \( x \):\[ 2x^5 + 4x^3b^2 - 16xb^4 = 0 \].Factor out \( 2x \) from the equation:\[ 2x(x^4 + 2x^2b^2 - 8b^4) = 0 \].Thus, the critical points are at \( x = 0 \) or when \( x^4 + 2x^2b^2 - 8b^4 = 0 \).

Step 5: Solve for x in the quadratic equation

Since \( x = 0 \) is not in the domain (as \( x > b \)), solve \( x^4 + 2x^2b^2 - 8b^4 = 0 \). Let \( u = x^2 \), then:\[ u^2 + 2ub^2 - 8b^4 = 0 \].Using the quadratic formula, \( u = \frac{-b^2 \pm \sqrt{b^4 + 8b^4}}{2} \), simplify and solve:\[ u = \frac{-b^2 \pm \sqrt{9b^4}}{2} \].\( u = b^2 \text{ or } -4b^2 \). Since \( u = x^2 > 0 \), \( x^2 = b^2 \), hence \( x = b \) is the only critical point, but only for theoretical analysis as it must be outside the domain.

Step 6: Apply the First Derivative Test

Critical points within the domain must be checked using \( f'(x) \) or any points returned from analysis if not crossing domain boundaries. For practical solving, repeat calculated critical numbers must satisfy domain without boundary issues. Plug back into derivative function to check intervals to understand increase/decrease change: Not needed yet since \( x = b \) intersects hypothetical boundary. Final practical evaluation must ensure given criteria fit for increasing or decreasing behavior not clearly seen without other opportunities.

Key Concepts

Critical Points

Critical points of a function are where its derivative is zero or undefined. These are significant because they help identify where a function might have a local maximum, minimum, or a saddle point.

To find critical points for our function \( f(x) = \frac{x^4 + 8b^4}{x^2 + b^2} \), we compute the derivative and set it to zero:

• Solve the equation \( 2x(x^4 + 2x^2b^2 - 8b^4) = 0 \).
• This gives \( x = 0 \) or the roots of the polynomial \( x^4 + 2x^2b^2 - 8b^4 = 0 \).

However, some solutions might not be valid if they are outside the domain of the function. Make sure to check if the critical points lie within the domain \( x > b \).

In our case, \( x = 0 \) is not in the domain, so it’s not a valid critical point. Solving \( x^4 + 2x^2b^2 - 8b^4 = 0 \) further gives \( x = b \), which is on the boundary of our domain. It requires careful interpretation and possibly needs reevaluation with respect to the function's domain constraints.

First Derivative Test

The First Derivative Test is a useful tool to determine whether a critical point is a local maximum, local minimum, or neither. This test relies on the behavior of the derivative before and after the critical point.

For a critical point at \( x = c \):

• If \( f'(x) \) changes from positive to negative as \( x \) crosses \( c \), then \( f \) has a local maximum at \( c \).
• If \( f'(x) \) changes from negative to positive, then \( f \) has a local minimum at \( c \).
• If \( f'(x) \) does not change signs, then \( f \) does not have a local extremum at \( c \).

In our exercise, since \( x = b \) is at the very edge of the domain \( x > b \), you might need to consider potential behavior immediately to the right of \( b \) or use limits. Analyze derivative values around these areas to decide the nature of these points if feasible.

It’s important to ensure all examined points are strictly within the domain to accurately apply the test and assess local behavior.

Function Domain

The function domain refers to the set of all possible input values (\( x \) values) for which the function is defined. For many rational functions, the domain is influenced by where the denominator is not zero.

In our problem, the function is:

• \( f(x) = \frac{x^4 + 8b^4}{x^2 + b^2} \)

Here, \( x^2 + b^2 eq 0 \) is always true because both terms are positive for real \( x \) and \( b > 0 \). The constraint \( b < x \) is hence the only domain restriction. This means the domain of \( f(x) \) is \( x > b \).

Understanding the domain is crucial for determining valid critical and test points when conducting calculus-based evaluations. Evaluating whether critical points are inside or outside the domain tells you whether they're functionally relevant or just theoretical.

Quotient Rule

The quotient rule is a technique used to take the derivative of a function that is the ratio of two other functions. It's applied when differentiating a function in the form \( \frac{g(x)}{h(x)} \).

The quotient rule formula is:

• \( f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2} \)

For the given function, define:

• \( g(x) = x^4 + 8b^4 \)
• \( h(x) = x^2 + b^2 \)

You would then need \( g'(x) \ = 4x^3 \) and \( h'(x) = 2x \). The derivative \( f'(x) \) will be computed using these results. Proceed step by step to precisely simplify and handle algebra.

The quotient rule is essential in calculus as many functions can be represented as ratios. Evaluating the function's derivative specifically in this exercise informs us about the potential presence of critical points.