Problem 24 - 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 \(1-24\). Your sketch should exhibit, and have labeled, all of the following: a) local and global extrema, b) inflection points, c) intervals on which function is increasing or decreasing, d) intervals on which function is concave up or concave down, e) horizontal and vertical asymptotes. $$ f(x)=1 / x-1 / x^{2} $$

Step-by-Step Solution

Answer

The function has a local maximum at \( x = 2 \) and an inflection point at \( x = 3 \). It increases on \( (0, 2) \) and decreases on \( (2, \infty) \). It is concave down on \( (0, 3) \) and concave up on \( (3, \infty) \). Vertical asymptote at \( x = 0 \), horizontal asymptote at \( y = 0 \).

Step 1: Identify Critical Points

To find local and global extrema, we first need the derivative of the function for critical points. The function is \( f(x) = \frac{1}{x} - \frac{1}{x^2} \). Differentiating gives \( f'(x) = -\frac{1}{x^2} + \frac{2}{x^3} = \frac{-x + 2}{x^3} \). Setting \( f'(x) = 0 \) yields the critical point at \( x = 2 \). Evaluate \( f(x) \) there to get \( f(2) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \).

Step 2: Determine Extrema Nature

Check the nature of the critical points using the first derivative test. For \( x < 2 \), \( f'(x) > 0 \) (positive), which means the function is increasing. For \( x > 2 \), \( f'(x) < 0 \) (negative), indicating that the function is decreasing. Therefore, \( x = 2 \) is a local maximum.

Step 3: Find Points of Inflection

Determine the second derivative \( f''(x) \) for inflection points. The second derivative \( f''(x) = \frac{2}{x^3} - \frac{6}{x^4} \) simplifies to \( \frac{2(x - 3)}{x^4} \). Setting \( f''(x) = 0 \) gives \( x = 3 \). Evaluate at \( x = 3 \) for inflection: \( f(3) = \frac{1}{3} - \frac{1}{9} = \frac{2}{9} \).

Step 4: Determine Function Intervals for Increasing/Decreasing

From the first derivative analysis: the function is increasing on \( (0, 2) \) because \( f'(x) > 0 \) and decreasing on \( (2, \infty) \) since \( f'(x) < 0 \).

Step 5: Determine Concavity Intervals

Using the second derivative: \( f''(x) > 0 \) when \( x > 3 \), indicating the function is concave up. If \( x < 3 \) but \( x > 0 \), \( f''(x) < 0 \), making the function concave down.

Step 6: Determine Asymptotes

Identify vertical and horizontal asymptotes. Vertical asymptote is at \( x = 0 \) because the function is undefined at \( x = 0 \). As \( x \to \infty \) or \( x \to -\infty \), \( f(x) \to 0 \) so the horizontal asymptote is \( y = 0 \).

Step 7: Sketch the Graph

Based on steps 1 to 6, sketch the graph. Label critical points at \( x = 2 \), inflection at \( x = 3 \), vertical asymptote at \( x = 0 \), horizontal asymptote at \( y = 0 \), and concavity changes between \( (0, 3) \) and \( (3, \infty) \).

Key Concepts

Critical Points

In calculus, critical points are where the function's slope is zero or undefined. For a function \( f(x) \), these are the points where \( f'(x) = 0 \) holds true or \( f'(x) \) doesn't exist. They can signal potential local maximums, minimums, or saddle points. In our function \( f(x) = \frac{1}{x} - \frac{1}{x^2} \), taking the derivative gives us \( f'(x) = \frac{-x + 2}{x^3} \).
Setting this derivative to zero, \( -x + 2 = 0 \), we find a critical point at \( x = 2 \). To determine the nature (whether it's a local max, min, or saddle), we use the first derivative test. If the function changes from increasing to decreasing at this point, it indicates a local maximum. Therefore, at \( x = 2 \) the derivative changes from positive to negative, confirming a local maximum at \( x = 2 \) with the function value \( f(2) = \frac{1}{4} \).

Derivative Analysis

Derivative analysis involves using the first derivative \( f'(x) \) to discover critical points and determine intervals where the function is increasing or decreasing. Referring to our function's derivative, \( f'(x) = \frac{-x + 2}{x^3} \), we already know that it's zero at \( x = 2 \). This analysis helps us spot behaviors of the function near these points.
For \( x < 2 \), the derivative is positive because \( f'(x) > 0 \), indicating the function increases over that interval. For \( x > 2 \), the derivative becomes negative \( f'(x) < 0 \), showing the function decreases. Understanding these variations aids in revealing the overall shape of the graph.

Concavity

Concavity refers to whether a function's graph bends upwards or downwards. This is determined through the second derivative \( f''(x) \). If \( f''(x) \) is positive, the function is concave up; if negative, concave down. This function’s second derivative is \( f''(x) = \frac{2(x - 3)}{x^4} \).
Setting \( f''(x) = 0 \) gives the inflection point at \( x = 3 \), where the concavity changes. An inflection point is where the graph changes from concave up to concave down or vice versa. For \( x > 3 \), \( f''(x) > 0 \) indicating concave up, while for \( x < 3 \), \( 0 < x < 3 \), \( f''(x) < 0 \), meaning concave down. Recognizing these intervals is crucial for sketching accurate graphs.

Asymptotes in Functions

Asymptotes are lines that a graph approaches but never touches or crosses. Vertical asymptotes occur where a function is undefined, while horizontal ones indicate the behavior as \( x \) heads to infinity. For our function, there’s a vertical asymptote at \( x = 0 \) since \( f(x) \) becomes undefined.
The horizontal asymptote of the function is \( y = 0 \), which is the value \( f(x) \) approaches as \( x \) heads towards infinity or negative infinity. This shows that despite fluctuations, the function's value gets arbitrarily close to zero. Understanding these asymptotes guides in predicting the behavior of a graph at its extremes and ensuring an accurate sketch reflects infinite behavior.

• Vertical at \( x = 0 \)
• Horizontal at \( y = 0 \)