Problem 48 - Calculus Single Variable

Question

Follow the outline given in this section to give a careful sketch of the graph of each of the functions in Exercises \(45-54 .\) Your sketch should exhibit, and have labeled, all of the following: (i) all local and global extrema, (ii) inflection points, (iii) intervals on which function is increasing or decreasing, (iv) intervals on which function is concave up or concave down, (v) all horizontal, vertical, and skewasymptotes. $$ f(x)=\frac{(x-1)(x-4)}{(x-3)(x-2)} $$

Step-by-Step Solution

Answer

The function has vertical asymptotes at \(x = 2\) and \(x = 3\), a horizontal asymptote at \(y = 1\), a local minimum at \(x = \frac{7}{3}\), and an inflection point at \(x = \frac{5}{2}\).

Step 1: Identify the Domain

Determine where the function is undefined. The function is undefined where the denominator is zero, i.e., when \((x-3)(x-2) = 0\). Thus, \(x = 3\) and \(x = 2\) are points where the function is undefined.

Step 2: Locate Vertical Asymptotes

Vertical asymptotes occur at values that make the denominator zero, where the numerator is non-zero. Here, as \(x\) approaches \(2\) or \(3\), the denominator becomes zero while the numerator does not, so \(x = 2\) and \(x = 3\) are vertical asymptotes.

Step 3: Find Horizontal Asymptotes

To find horizontal asymptotes, compare the degrees of the polynomials in the numerator and denominator. Both are degree 2, so the horizontal asymptote is given by the ratio of the leading coefficients, \(\frac{1}{1} = 1\). Thus, \(y = 1\) is a horizontal asymptote.

Step 4: Calculate Critical Points

Find the derivative \(f'(x)\) and set it to zero to find critical points. Use the quotient rule for differentiation. After simplification, solve \(f'(x) = 0\) for \(x\) to find critical points: \(x = \frac{7}{3}\). Also, check points \(x = 2\) and \(x = 3\) though these are undefined, so no extrema here.

Step 5: Determine Intervals of Increase/Decrease

Analyze the sign of \(f'(x)\) in three intervals: \((-\infty, 2)\), \((2, 3)\), and \((3, \infty)\). Based on the sign of \(f'(x)\), the function is decreasing on \((-\infty, 2)\), increasing on \((2, 3)\), and decreasing on \((3, \infty)\).

Step 6: Identify Local/Global Extrema

Using intervals from Step 5, determine that \(x = \frac{7}{3}\) is a local minimum because the function changes from decreasing to increasing at this point.

Step 7: Find Inflection Points

Compute \(f''(x)\) using the derivative from Step 4. Set \(f''(x) = 0\) and solve for \(x\) to find potential inflection points. Verify by checking the change in concavity, resulting in points such as \(x = \frac{5}{2}\).

Step 8: Intervals of Concavity

Check the sign of \(f''(x)\) over the intervals split by the second derivative's critical points: \((-\infty, \frac{5}{2})\) and \((\frac{5}{2}, \infty)\). The function is concave down on \((-\infty, \frac{5}{2})\) and concave up on \((\frac{5}{2}, \infty)\).

Step 9: Sketch the Graph

Starting from all gathered information, sketch the function \(f(x)\), marking vertical asymptotes at \(x = 2, 3\), the horizontal asymptote \(y = 1\), the local minimum at \(x = \frac{7}{3}\), and the inflection point at \(x = \frac{5}{2}\). The sketch should show the function's increasing and decreasing behavior, as well as intervals of concavity.

Key Concepts

Asymptotes

In graphing functions, asymptotes are critical as they indicate lines that the curve approaches but never actually touches. For the function \(f(x) = \frac{(x-1)(x-4)}{(x-3)(x-2)}\), we have two types of asymptotes to consider: vertical and horizontal.

• Vertical Asymptotes: These occur where the function is undefined. For our function, vertical asymptotes are found by setting the denominator equal to zero. When \((x-3)(x-2) = 0\), we find asymptotes at \(x = 2\) and \(x = 3\).
• Horizontal Asymptotes: These are revealed by comparing the degrees of the numerator and denominator. Since both have a degree of 2, the horizontal asymptote is determined by the leading coefficients. Thus, \(y = 1\) is a horizontal asymptote.

Understanding and sketching these asymptotes on the graph provides a framework for plotting the function.

Critical Points

Critical points of a function are where its derivative equals zero or is undefined; these can be potential locations for local maxima or minima. To find these for \(f(x)\), we calculate its derivative using the quotient rule.

• The derivative \(f'(x)\) is derived and set to zero to detect critical points of interest. In this exercise, solving \(f'(x) = 0\) gives us \(x = \frac{7}{3}\).
• While critical points often indicate local minima or maxima, they must first be tested. The local nature is verified by examining the change in behavior of the function around these points.

This process helps in determining critical changes in the graph's slope, signaling a potential shift in increasing or decreasing trends.

Intervals of Concavity

Intervals of concavity showcase where a graph is curving upward or downward. To find these, the second derivative \(f''(x)\) of \(f(x)\) must be utilized.

• To determine intervals of concavity, the sign of \(f''(x)\) is analyzed over various intervals determined by inflection points. In this function, the concavity changes at \(x = \frac{5}{2}\).
• The notion of concavity encompasses two possibilities:
  • Concave Up: Occurs when \(f''(x) > 0\).
  • Concave Down: Occurs when \(f''(x) < 0\).
  For \(f(x)\), we find the function is concave down from \((-\infty, \frac{5}{2})\) and concave up on \((\frac{5}{2}, \infty)\).

This understanding helps in rounding out how the graph is shaped beyond just the linear direction it takes.

Inflection Points

Inflection points are where the concavity of a function changes, offering additional insight into the graph's behavior, particularly how it changes curvature.

• Determining Inflection Points: These are discovered by setting the second derivative \(f''(x)\) to zero. In the given function \(f(x)\), solving \(f''(x) = 0\) pinpoints a key inflection point at \(x = \frac{5}{2}\).
• Concavity Examination: At an inflection point, the function switches between concave down and concave up or vice-versa, highlighting a significant change in the graph's curvature.

Locating inflection points enables an accurate sketch of the function by showing where the curve itself changes direction across the axes.