Problem 43 - Calculus with Concepts in Calculus

Question

Use the formulas for the function and its first and second derivatives as an aid in sketching the graph of the given function. Note all relevant properties listed in Table 4.1. $$ \begin{aligned} &f(x)=\frac{x}{\sqrt{1-x}}, f^{\prime}(x)=\frac{2-x}{2(1-x)^{3 / 2}}, \text { and } \\ &f^{\prime \prime}(x)=\frac{4-x}{4(1-x)^{5 / 2}} \end{aligned} $$

Step-by-Step Solution

Answer

Function \( f(x) \) is increasing on \( (-\infty, 2) \), decreasing on \( (2, 1) \), has vertical asymptote at \( x = 1 \), and is concave up with local maximum at \( x = 2 \).

Step 1: Determine Domain

The function \( f(x) = \frac{x}{\sqrt{1-x}} \) has a square root in the denominator, which implies that the expression inside the square root must be positive. Thus, the domain of \( f(x) \) is \( x < 1 \). Additionally, to avoid division by zero, \( x eq 1 \). Hence, the domain of \( f(x) \) is \( (-\infty, 1) \).

Step 2: Find Critical Points

Critical points occur where the derivative is zero or undefined. The derivative \( f'(x) = \frac{2-x}{2(1-x)^{3/2}} \) is undefined at \( x = 1 \). Setting the numerator \( 2-x \) equal to zero gives \( x = 2 \). Thus, \( x = 2 \) is a critical point within the domain.

Step 3: Analyze First Derivative for Monotonicity

Using \( f'(x) = \frac{2-x}{2(1-x)^{3/2}} \), determine sign changes. For \( x < 1 \), the expression \( (1-x)^{3/2} \) is positive. Thus, \( f'(x) > 0 \) when \( x < 2 \) and \( f'(x) < 0 \) when \( x > 2 \). This indicates \( f(x) \) is increasing on \( (-\infty, 2) \) and decreasing on \( (2, 1) \).

Step 4: Identify Inflection Points Using Second Derivative

Find where the second derivative \( f''(x) = \frac{4-x}{4(1-x)^{5/2}} \) is zero or changes sign. Setting \( 4-x = 0 \) gives \( x = 4 \). However, \( x = 4 \) is outside the domain \( (-\infty, 1) \), so there are no inflection points within the domain. Analyze sign changes: \( f''(x) > 0 \) for \( x < 4 \), indicating the graph is concave up on \( (-\infty, 1) \).

Step 5: Note Asymptotic Behavior

As \( x \to 1 \) from the left, the denominator \( \sqrt{1-x} \to 0 \) and \( f(x) \to \infty \), indicating a vertical asymptote at \( x = 1 \). As \( x \to -\infty \), \( f(x) \to 0 \), suggesting a horizontal asymptote at \( y = 0 \).

Step 6: Sketch the Graph

Combine the information: increasing on \( (-\infty, 2) \), decreasing on \( (2, 1) \), vertical asymptote at \( x = 1 \), horizontal asymptote at \( y = 0 \), and concave up on \( (-\infty, 1) \). The critical point at \( x = 2 \) is a local maximum.

Key Concepts

Domain of Function

When dealing with functions, identifying the domain is a fundamental step in graph sketching. The domain represents all possible input values (x-values) that do not cause the function to become undefined.
For the function given, \( f(x) = \frac{x}{\sqrt{1-x}} \), the expression within the square root, \( 1-x \), must be greater than zero to avoid taking the square root of a negative number. Thus, \( x < 1 \), ensuring the denominator doesn't become zero. Additionally, since we have a square root in the denominator, \( x eq 1 \) to prevent division by zero.

• Therefore, the domain of \( f(x) \) is all real numbers less than 1, or \( (-\infty, 1) \).

Critical Points

Critical points of a function indicate where the function's graph changes direction or has a horizontal tangent line. These points occur where the derivative is zero or undefined.
For the function \( f(x) \), the first derivative is given by \( f'(x) = \frac{2-x}{2(1-x)^{3/2}} \). It is undefined at \( x = 1 \), which is outside the domain, so we ignore this point.

• Setting the numerator \( 2-x = 0 \) gives the critical point \( x = 2 \), which is important to note within the domain \( (-\infty, 1) \).

First and Second Derivatives

The first derivative, \( f'(x) \), helps determine where the function is increasing or decreasing, thereby indicating monotonicity. For the given function, it shows us:

• \( f'(x) > 0 \) when \( x < 2 \), meaning the function is increasing.
• \( f'(x) < 0 \) when \( x > 2 \), indicating the function is decreasing.

This analysis shows that \( f(x) \) increases on \( (-\infty, 2) \) and decreases on \( (2, 1) \), making \( x = 2 \) a local maximum.
The second derivative \( f''(x) = \frac{4-x}{4(1-x)^{5/2}} \) suggests concavity of the function graph. A change in the sign of the second derivative typically indicates an inflection point, but \( x = 4 \), where the second derivative zeroes, is outside the domain.

• For \( x < 4 \), \( f''(x) > 0 \), meaning the graph is concave up within its domain.

Asymptotic Behavior

Asymptotic behavior pertains to the trends of the graph as it approaches certain lines. These are crucial for understanding long-term behavior of functions.
For the given function \( f(x) \), as \( x \) approaches 1 from the left, \( \sqrt{1-x} \) tends to zero. Thus, \( f(x) \to \infty \) shows a vertical asymptote at \( x = 1 \).
When considering the limits as \( x \to -\infty \), \( f(x) \to 0 \), indicating a horizontal asymptote at \( y = 0 \).

• These asymptotes guide us when sketching the graph, revealing the vertical stretch near \( x = 1 \) and the horizontal stretch as \( x \to -\infty \).