Problem 22 - 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. $$ f(x)=x+54 / \sqrt{x} $$

Step-by-Step Solution

Answer

The function \( f(x) \) has a local minimum at \( (9, 27) \), is decreasing on \( (0, 9) \), increasing on \( (9, \infty) \), and is concave up everywhere with no asymptotes.

Step 1: Determine Domain

The function is given as \( f(x) = x + \frac{54}{\sqrt{x}} \). The square root in the denominator implies that \( x > 0 \) because the square root of a negative number is undefined in the real number system, and the expression cannot equal zero in the denominator. Therefore, the domain of \( f(x) \) is \( (0, \infty) \).

Step 2: Find Critical Points

First, find the derivative of \( f(x) \): \( f'(x) = 1 - \frac{27}{x^{3/2}} \). Set this derivative to zero to find critical points: \( 1 - \frac{27}{x^{3/2}} = 0 \). Solving for \( x \), we get \( x^{3/2} = 27 \), hence \( x = 9 \). So, \( x = 9 \) is a critical point.

Step 3: Determine Increasing/Decreasing Intervals

Using the critical point from Step 2, determine where the function increases or decreases by evaluating the sign of \( f'(x) \) on the intervals \((0, 9)\) and \((9, \infty)\). For \( x \in (0, 9) \), \( f'(x) < 0 \), so \( f(x) \) is decreasing. For \( x \in (9, \infty) \), \( f'(x) > 0 \), so \( f(x) \) is increasing.

Step 4: Identify Extrema

Since \( f(x) \) transitions from decreasing to increasing at \( x = 9 \), \( f(x) \) has a local minimum at \( x = 9 \). Calculate \( f(9) = 9 + \frac{54}{3} = 27 \), so the local minimum is \( (9, 27) \). There is no global maximum or minimum since neither of the endpoints are included in the domain.

Step 5: Find Inflection Points

To locate inflection points, find the second derivative: \( f''(x) = \frac{81}{2x^{5/2}} \). This expression is positive for all \( x > 0 \), so \( f(x) \) is always concave up and has no inflection points.

Step 6: Asymptotic Behavior

Consider horizontal asymptotes as \( x \to \infty \): \( f(x) = x + \frac{54}{\sqrt{x}} \) approximates \( x \), so \( f(x) \to \infty \). For vertical behavior as \( x \to 0^+ \), \( \frac{54}{\sqrt{x}} \to \infty \), so \( f(x) \to \infty \). Therefore, there are no asymptotes.

Step 7: Sketch the Graph

Based on the analysis, draw the graph: start decreasing from an undefined \(x \to 0^+\), reaching a minimum at \( (9, 27) \), and increasing thereafter. The graph should not have any asymptotes, nor any change in concavity as determined.

Key Concepts

Critical Points

In graph sketching, identifying critical points is essential to understanding where a function might change its behavior in terms of increasing or decreasing trends. A critical point occurs where the derivative of the function equals zero or where it is undefined. For the function given, \( f(x) = x + \frac{54}{\sqrt{x}} \), we derive it to find: \( f'(x) = 1 - \frac{27}{x^{3/2}} \).

To find the critical points, we set the derivative equal to zero: \( 1 - \frac{27}{x^{3/2}} = 0 \). Solving this equation results in \( x = 9 \). Thus, \( x = 9 \) is identified as a critical point.

Critical points help us locate local minimums and maximums, where the function potentially "turns" to change its direction. Understanding these concepts enables a precise graph sketch that accurately represents the function's behavior over its domain.

Increasing/Decreasing Intervals

Once the critical points are identified, they can be used to define intervals on which the function is either increasing or decreasing. With the critical point at \( x = 9 \) from the derivative \( f'(x) = 1 - \frac{27}{x^{3/2}} \), we determine the intervals by checking the sign of the derivative.

- For \( x \in (0, 9) \), the derivative \( f'(x) < 0 \) indicates that the function is decreasing.- For \( x \in (9, \infty) \), the derivative \( f'(x) > 0 \) suggests the function is increasing.

These intervals provide insight into the behavior of the function across different sections of the graph. The transition from decreasing to increasing at \( x = 9 \) not only confirms \( x = 9 \) as a local minimum but shows the change in trend that shapes the graph's slope.

Concavity

In graph sketching, understanding concavity is crucial since it reveals how the curve opens—whether it bends upwards or downwards. This is determined by the second derivative of the function. For \( f(x) = x + \frac{54}{\sqrt{x}} \), the second derivative is calculated as:

\[ f''(x) = \frac{81}{2x^{5/2}} \]
The positive value of \( f''(x) \) for all \( x > 0 \) indicates the function is concave up throughout its domain. Since a concave up shape resembles a "U," the curve opens upwards but not sharply, reflecting a gentle, upward curvature.

Concavity analysis can also highlight possible points of inflection, where the concavity changes sign. In this case, since the sign does not change, there are no inflection points confirming the constant upward bend.

Asymptotic Behavior

Asymptotic behavior examines how a function behaves as it approaches the boundaries of its domain. Particularly, we look at horizontal and vertical asymptotes to predict this behavior.

For the function \( f(x) = x + \frac{54}{\sqrt{x}} \):

• As \( x \to \infty \), the term \( \frac{54}{\sqrt{x}} \to 0 \), thus \( f(x) \to x \). This suggests the function continues to rise, approximating an upward linear trend without reaching a horizontal asymptote.
• As \( x \to 0^+ \), \( \frac{54}{\sqrt{x}} \to \infty \), hence \( f(x) \to \infty \). Therefore, no vertical asymptote exists since it suggests an upward trend surpassing any finite boundary near the origin.

Understanding asymptotic behavior allows us to anticipate the function's boundaries and how it proceeds to infinity, confirming the absence of both horizontal and vertical asymptotes in this scenario. This insight aids in accurately sketching the graph by indicating the trends beyond just critical points and intervals.