Problem 4 - Active Calculus

Question

Let \(q(x)=\frac{e^{-x}}{x-c}\) be a one-parameter family of functions where \(c>0\). a. Explain why \(q\) has a vertical asymptote at \(x=c\). b. Determine \(\lim _{x \rightarrow \infty} q(x)\) and \(\lim _{x \rightarrow-\infty} q(x)\). c. Compute \(q^{\prime}(x)\) and find all critical numbers of \(q\). d. Construct a first derivative sign chart for \(q\) and determine whether each critical number leads to a local minimum, local maximum, or neither for the function \(q\). e. Sketch a typical member of this family of functions with important behaviors clearly labeled.

Step-by-Step Solution

Answer

a. Vertical asymptote at \( x = c \).b. \( \lim_{x \rightarrow \infty} q(x) = 0 \), \( \lim_{x \rightarrow -\infty} q(x) = -\infty \).c. \( q'(x) = \frac{e^{-x}(x - c + 1)}{(x-c)^2} \), critical number at \( x = c - 1 \).

Step 1: Identify where the vertical asymptote occurs

The function has a vertical asymptote where the denominator is zero. For the function \( q(x)=\frac{e^{-x}}{x-c} \), the denominator becomes zero when \( x = c \). As a result, the function \( q(x) \) has a vertical asymptote at \( x = c \).

Step 2: Calculate the limit as x approaches infinity

For the limit as \( x \) approaches infinity, consider \[ q(x) = \frac{e^{-x}}{x-c} \]. As \( x \to \, \infty \), \(e^{-x} \) approaches 0 and \( x-c \) approaches infinity. Therefore, \[ \lim_{x \to \, \infty} \frac{e^{-x}}{x-c} = 0 \].

Step 3: Calculate the limit as x approaches negative infinity

For the limit as \( x \) approaches negative infinity, consider \[ q(x) = \frac{e^{-x}}{x-c} \]. As \( x \to \, -\infty \), \(e^{-x} \) approaches infinity and \( x-c \) approaches negative infinity. Therefore, \[ \lim_{x \to \, -\infty} \frac{e^{-x}}{x-c} = -\infty \].

Step 4: Compute the derivative of q(x)

To find the critical numbers, first compute the derivative of \( q(x) \). Using the quotient rule with \( u = e^{-x} \) and \( v = x - c \), where \( u' = -e^{-x} \) and \( v' = 1 \), we get: \[ q'(x) = \frac{(e^{-x})(1) - (-e^{-x})(x-c)}{(x-c)^2} = \frac{e^{-x} + e^{-x}(x-c)}{(x-c)^2} = \frac{e^{-x}(x -c + 1)}{(x-c)^2} \].

Step 5: Find critical numbers

Critical numbers occur where the first derivative is equal to zero or undefined. \( q'(x) = 0 \) implies: \[ \frac{e^{-x}(x-c+1)}{(x-c)^2} = 0 \]. Setting the numerator equal to zero gives: \[ e^{-x}(x-c+1) = 0 \]. Since \( e^{-x} eq 0 \), it follows that: \[ x - c + 1 = 0 \implies x = c - 1 \]. Hence, the critical number is \( x = c - 1 \).

Step 6: Construct the first derivative sign chart

Determine the sign of \( q'(x) \) around the critical number \( x = c - 1 \) and the vertical asymptote \( x = c \). Test intervals such as \( x < c - 1 \), \( c - 1 < x < c \), and \( x > c \): 1. For \( x < c - 1 \), both \( e^{-x} \) and \( x - c + 1 \) are positive, making \( q'(x) > 0 \). 2. For \( c - 1 < x < c \), \( e^{-x} \) is positive, but \( x - c + 1 \) is negative, making \( q'(x) < 0 \). 3. For \( x > c \), both \( e^{-x} \) and \( x - c + 1 \) are positive, making \( q'(x) > 0 \). Therefore, \( q(x) \) has a local maximum at \( x = c - 1 \).

Step 7: Sketch the function

Based on the information above: - The function \( q(x) \) has a vertical asymptote at \( x = c \). - It approaches 0 as \( x \) approaches \( +\infty \). - It approaches \( - \infty \) as \( x \) approaches \( - \infty \). - It has a local maximum at \( x = c - 1 \). Sketch these features to display the function's behavior.

Key Concepts

vertical asymptote

A vertical asymptote is a vertical line where a function's value becomes unbounded or undefined. In the given function, \(q(x) = \frac{e^{-x}}{x-c}\), a vertical asymptote occurs when the denominator is zero. Therefore, setting \(x - c = 0\) gives us \(x = c\). This means that as \(x\) approaches \(c\), the value of \(q(x)\) grows without limit, causing the graph to approach a line at \(x = c\). This is typical behavior in rational functions when the denominator is zero, and it prevents the function from being defined at that point.

limits at infinity

Limits at infinity describe the behavior of a function as \(x\) approaches positive or negative infinity. For \(q(x) = \frac{e^{-x}}{x-c}\):

• As \(x\) approaches positive infinity (\(x \rightarrow +\infty\)), \(e^{-x}\) rapidly approaches zero because the exponent is very large and negative. Meanwhile, \(x - c\) increases, making the fraction approach zero.
• As \(x\) approaches negative infinity (\(x \rightarrow -\infty\)), \(e^{-x}\) becomes exponentially large because negative and smaller \(x\) values make the exponent positive. \(x-c\) becomes a large negative number, leading the fraction to approach negative infinity.

Therefore, \

• \(\lim_{x \to \infty} \frac{e^{-x}}{x-c} = 0\)
• \(\lim_{x \to -\infty} \frac{e^{-x}}{x-c} = -\infty\)

derivative

The derivative represents how a function changes as its input changes. To find the derivative \(q'(x)\) of \(q(x) = \frac{e^{-x}}{x-c}\), we use the quotient rule: \

• Let \(u = e^{-x}\) and \(v = x - c\)
• Then \(u' = -e^{-x}\) and \(v' = 1\)
• The quotient rule states \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\)
• Applying this, we get \(q'(x) = \frac{e^{-x}(1) - (-e^{-x})(x-c)}{(x-c)^2} = \frac{e^{-x} + e^{-x}(x-c)}{(x-c)^2} = \frac{e^{-x}(x - c + 1)}{(x-c)^2}\)

This derivative helps us analyze how the function \(q(x)\) behaves, whether it's increasing or decreasing, which is critical in finding maximum and minimum points.

critical numbers

Critical numbers are the values of \(x\) where the first derivative \(q'(x)\) is zero or undefined. Critical numbers indicate where a function might have local maxima, minima, or saddle points. To find the critical numbers of \(q(x) = \frac{e^{-x}}{x-c}\), we set \(q'(x) = 0\): \

• \(\frac{e^{-x}(x - c + 1)}{(x-c)^2} = 0\)
• The numerator must be zero: \(e^{-x}(x - c + 1) = 0\)
• Since \(e^{-x} eq 0\), we solve \(x - c + 1 = 0 \Rightarrow x = c - 1\)

Thus, the critical number for the function \(q(x)\) is \(x = c - 1\). This helps in identifying points where the function changes direction.

first derivative test

The first derivative test helps determine whether a critical number corresponds to a local maximum or minimum. After calculating the first derivative \(q'(x) = \frac{e^{-x}(x - c + 1)}{(x-c)^2}\), we analyze its sign around the critical number \(x = c - 1\):

• For \(x < c - 1\): Both \(e^{-x}\) and \(x - c + 1\) are positive, making \(q'(x) > 0\). The function is increasing.
• For \(c - 1 < x < c\): \(e^{-x}\) is positive, but \(x - c + 1\) is negative, making \(q'(x) < 0\). The function is decreasing.
• For \(x > c\): Both \(e^{-x}\) and \(x - c + 1\) are positive, making \(q'(x) > 0\). The function is increasing again.

The sign changes around \(x = c - 1\), indicating a local maximum at this point. The test provides a clear method to determine the nature of critical points, aiding in sketching the function's graph.