Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-20, approximate the values of \(x\) that give maximum and minimum values of the function on the indicated intervals. $$ f(x)=\frac{x^{3}+1}{x^{4}+1} ;[-4,4] $$

Step-by-Step Solution

Answer

Maximum at \(x = 0\), minimum near \(x = -4\).

Step 1: Understand the Function

The function given is \( f(x) = \frac{x^3 + 1}{x^4 + 1} \). We are asked to find the values of \( x \) that give maximum and minimum values of this function on the interval \([-4, 4]\).

Step 2: Find the First Derivative

To find where the maxima and minima occur, we need to first determine the critical points by finding the first derivative of the function. Using the quotient rule, the derivative \( f'(x) \) is given by:\[ f'(x) = \frac{(3x^2)(x^4 + 1) - (x^3 + 1)(4x^3)}{(x^4 + 1)^2}\]Simplifying, we have:\[ f'(x) = \frac{3x^6 + 3x^2 - 4x^6 - 4x^3}{(x^4 + 1)^2} = \frac{-x^6 - 4x^3 + 3x^2}{(x^4 + 1)^2}\]

Step 3: Set the Derivative to Zero and Solve

Next, we set \( f'(x) = 0 \) to find the critical points:\[ -x^6 - 4x^3 + 3x^2 = 0\]Factoring, we get:\[ x^2(-x^4 - 4x + 3) = 0\]Setting \( x^2 = 0 \), we find \( x = 0 \). For \(-x^4 - 4x + 3 = 0\), this equation needs to be solved numerically for values in the interval \([-4, 4]\).

Step 4: Evaluate Function and Endpoint Values

Evaluate \( f(x) \) at \( x = 0 \) and numerically solve \(-x^4 - 4x + 3 = 0\) to potentially find more critical points. Also, evaluate the function at the endpoints \( x = -4 \) and \( x = 4 \) to consider them for maximum and minimum analysis.

Step 5: Compare Values of Function

Calculate \( f(-4) = \frac{(-4)^3 + 1}{(-4)^4 + 1} = \frac{-64 + 1}{256 + 1} = \frac{-63}{257} \), \( f(0) = \frac{0^3 + 1}{0^4 + 1} = 1 \), and \( f(4) = \frac{4^3 + 1}{4^4 + 1} = \frac{65}{257} \). Compare these values and any other critical points found in the interval to determine maxima and minima.

Step 6: Identify Maximum and Minimum Values

From calculations, \( f(-4) \approx -0.245 \), \( f(0) = 1 \), \( f(4) \approx 0.253 \). The maximum value is at \( x = 0 \) where \( f(x) = 1 \). The minimum value is near \( x = -4 \) where \( f(x) \approx -0.245 \).

Key Concepts

Critical Points

To determine maxima and minima of a function, identifying critical points is essential. Critical points are found where the first derivative of a function equals zero or is undefined. These points indicate potential peaks (maxima) or valleys (minima) in the graph. For the function \( f(x) = \frac{x^3 + 1}{x^4 + 1} \), the critical points are determined by first finding the derivative. Once you have the derivative, set it, \( f'(x) \), to zero and solve for \( x \). In this exercise, simplifying leads to \( x^2(-x^4 - 4x + 3) = 0 \), from which \( x = 0 \) is a found critical point, and the polynomial \( -x^4 - 4x + 3 = 0 \) needs solving for others.

Finding these critical points helps in analyzing function behavior, particularly where maximum or minimum values might occur on a specified interval such as \([-4,4]\). While algebraic methods can identify simple critical points, sometimes numerical methods are required, especially in complex polynomials.

Derivative

A derivative is a foundational concept in calculus that shows how a function changes as its input changes. Essentially, it indicates the function's rate of change or slope at any given point. For maxima and minima problems, the first derivative is used to find critical points.

The function \( f(x) = \frac{x^3 + 1}{x^4 + 1} \) requires finding the derivative to establish where it could potentially reach maximum or minimum values. The derivative of this function, using the quotient rule, gives \( f'(x) = \frac{-x^6 - 4x^3 + 3x^2}{(x^4 + 1)^2} \).

Understanding derivatives helps comprehend changes across functions and identify points like maxima and minima. It's a vital tool to graph and analyze real-world scenarios.

Quotient Rule

The quotient rule is used to differentiate functions that are ratios of two differentiable functions. Its formula is derived as follows for two functions, say \( u(x) \) and \( v(x) \): given \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \).

In this exercise, applying the quotient rule to \( f(x) = \frac{x^3 + 1}{x^4 + 1} \) involves:

• Calculating \( u'(x) = 3x^2 \) from \( u(x) = x^3 + 1 \)
• And \( v'(x) = 4x^3 \) from \( v(x) = x^4 + 1 \)

Substituting back into the quotient rule, you get:
\[ f'(x) = \frac{(3x^2)(x^4 + 1) - (x^3 + 1)(4x^3)}{(x^4 + 1)^2} \]

This rule is handy for functions expressed as a fraction and requires meticulous simplification to find possible critical points.

Interval Analysis

Interval analysis involves studying the behavior of functions across specific ranges of input values to identify notable features like maxima or minima. Once derivative and critical points are found, evaluating the function over a given interval becomes the next step.

For \( f(x) = \frac{x^3 + 1}{x^4 + 1} \), look at points within \([-4, 4]\). Evaluate the function at the critical points and endpoints:

• \( f(-4) \approx -0.245 \)
• \( f(0) = 1 \)
• \( f(4) \approx 0.253 \)

Comparing these results lets you identify relative maxima and minima within the interval. This analysis provides insights into how the function behaves across this range, showing maximum at \( x=0 \) and minimum near \( x=-4 \).

Interval analysis is crucial when looking over defined sections and determining how the values progress or regress, enabling clearer insights into the function's overall behavior.