Problem 30 - Calculus Early Transcendentals

Question

Use a graphing utility to estimate the absolute maximum and minimum values of \(f \), if any, on the stated interval, and then use calculus methods to find the exact values. \(f(x)=(x-1)^{2}(x+2)^{2} ;(-\infty,+\infty)\)

Step-by-Step Solution

Answer

Maximum is \( \frac{81}{16} \) at \( x = -\frac{1}{2} \); minimum is 0 at \( x = 1 \) and \( x = -2 \).

Step 1: Understanding the Function

The given function is \(f(x) = (x-1)^2(x+2)^2\). It is a polynomial function, which means it is continuous over all real numbers.

Step 2: Use Graphing Utility

Use a graphing calculator or software to plot the function \(f(x) = (x-1)^2(x+2)^2\). Observe the graph to identify where the function appears to achieve its minimum and maximum values.

Step 3: Relate Observations to Calculus

From the graph, you might observe that the function appears to have a minimum at \(x = 1\) and a maximum at \(x = -2\). However, to confirm, we should calculate using calculus methods.

Step 4: Find Critical Points with Calculus

First, find the derivative of the function:\[ f'(x) = \frac{d}{dx}((x-1)^2(x+2)^2) \].\Apply the product rule: let \( u = (x-1)^2 \) and \( v = (x+2)^2 \), then \( f'(x) = u'v + uv' \).Calculate:\( u' = 2(x-1) \), \( v' = 2(x+2) \), so\[ f'(x) = 2(x-1)(x+2)^2 + (x-1)^2 2(x+2) \]Simplify and solve \( f'(x) = 0 \) for critical points.

Step 5: Solve the Derivative Equation

Simplify \( f'(x) \):\( f'(x) = 2(x-1)(x+2)^2 + 2(x-1)^2(x+2) \)= \( 2(x-1)(x+2)((x+2) + (x-1)) \)= \( 2(x-1)(x+2)(2x+1) \)Solve \( 2(x-1)(x+2)(2x+1) = 0 \). The critical points are where each factor separately equals zero: \( x = 1 \), \( x = -2 \), and \( 2x+1 = 0 \) which gives \( x = -\frac{1}{2} \).

Step 6: Evaluate Function at Critical Points

Plug the critical points back into the original function:- \( f(1) = (1-1)^2(1+2)^2 = 0 \)- \( f(-2) = (-2-1)^2(-2+2)^2 = 0 \)- \( f(-\frac{1}{2}) = (-\frac{1}{2}-1)^2(-\frac{1}{2}+2)^2 = \left(-\frac{3}{2}\right)^2 \left(\frac{3}{2}\right)^2 = \frac{81}{16} \)Thus, the maximum value is \( \frac{81}{16} \) and the minimum value is \( 0 \).

Step 7: Verify End Behavior

Since \( f(x) = (x-1)^2(x+2)^2 \) is a fourth-degree polynomial, its end behavior as \( x \to \pm \infty \) is \( f(x) \to +\infty \). Thus, the critical points found cover the possible extrema over all real numbers.

Key Concepts

Polynomial Function Analysis

Analyzing polynomial functions is an essential aspect of calculus. Polynomial functions are expressed as sums of terms consisting of a variable raised to an exponent and a coefficient. In this problem, the function is \( f(x) = (x - 1)^2(x + 2)^2 \), which is a fourth-degree polynomial. This implies that the polynomial has four roots, since the highest power of \( x \) is four. Polynomial functions are continuous and smooth; they have no breaks, holes, or sharp corners in their graphs. Because this is a fourth-degree polynomial, we know several things about its behavior:

• It behaves like \( x^4 \), meaning its end behavior as \( x \to \pm \infty \) is to approach \( +\infty \).
• Its roots are critical since they are potential points where the graph intersects the x-axis.

The function \( f(x) \), a product of squared binomials, may have distinct characteristics around its zeros that need closer examination to identify potential maxima or minima.

Critical Points

Identifying critical points in a function is crucial for determining where a function might achieve its highest or lowest values, known as extrema. Critical points occur where the derivative of a function equals zero or where the derivative is undefined. For the function \( f(x) = (x - 1)^2(x + 2)^2 \), we start by finding its derivative:To do this, apply the product rule since \( f(x) \) is the product of two separate binomial expressions:

• Let \( u = (x-1)^2 \)
• Let \( v = (x+2)^2 \)

Therefore, \( f'(x) = u'v + uv' \). Calculating these derivatives will help find where \( f'(x) = 0 \), revealing critical points. The process results in solving the equation \( 2(x-1)(x+2)(2x+1) = 0 \), yielding \( x = 1 \), \( x = -2 \), and \( x = -\frac{1}{2} \). These critical points indicate potential extrema of the original function.

Graphing Utilities

Graphing utilities are powerful tools used to visually represent functions and their behavior across different intervals. They help estimate places where a function achieves maximum or minimum values before precise calculus methods confirm these observations. For our function \( f(x) = (x - 1)^2(x + 2)^2 \), using a graphing calculator or software can provide insight into:

• The overall shape and symmetry of the graph: In this case, it shows the symmetry characteristic of fourth-degree polynomials.
• Where the function appears to have high and low points, hinting at possible extrema.

The graph reveals areas that might be maxima or minima, such as near the values \( x = -2 \) and \( x = 1 \). Though these provide preliminary estimates, it is crucial to supplement with calculus for exact results.

Derivative Calculation

The calculation of a derivative provides a formal mathematical way to determine the rate at which a function's value changes. It's especially useful for finding critical points, where the slope of the tangent is zero or undefined, indicating potential maxima or minima.To calculate the derivative of \( f(x) = (x-1)^2(x+2)^2 \), one must employ the product rule. This derivative, \( f'(x) = u'v + uv' \), simplifies to \( 2(x-1)(x+2)(2x+1) \).Key steps involve:

• Deriving each binomial separately: \( u' = 2(x-1) \) and \( v' = 2(x+2) \).
• Substituting these into the formula to find \( f'(x) \).
• Solving \( f'(x) = 0 \) to pinpoint critical points.

Understanding and correctly applying these derivative rules leads to determining exactly where a function like this one may achieve its extreme values.