Problem 42 - Calculus for Biology and Medicine

Question

Let $$ f(x)=\frac{2}{1+e^{-x}}, x \in \mathbf{R} $$ (a) Determine where \(f(x)\) is increasing and where it is decreasing. (b) Where is the function concave up and where is it concave down? Find all inflection points of \(f(x)\). (c) Find \(\lim _{x \rightarrow \infty} f(x)\) and decide whether \(f(x)\) has a horizontal asymptote. (d) Find \(\lim _{x \rightarrow-\infty} f(x)\) and decide whether \(f(x)\) has a horizontal asymptote. (e) Sketch the graph of \(f(x)\) together with its asymptotes and inflection points (if they exist).

Step-by-Step Solution

Answer

(a) Increasing for all \( x \); (b) Concave down on \( (-\infty, 0) \), concave up on \( (0, \infty) \), inflection at \( x = 0 \); (c) Horizontal asymptote at \( y = 2 \); (d) Horizontal asymptote at \( y = 0 \); (e) Sketch shows increasing function with asymptotes and inflection.

Step 1: Find the First Derivative

We start by finding the first derivative of the function \( f(x) = \frac{2}{1+e^{-x}} \). The derivative of \( f(x) \) can be found using the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). For our function, \( u = 2 \) and \( v = 1 + e^{-x} \). Find \( u' = 0 \) and \( v' = -e^{-x} \). Plug these into the formula to get:\[ f'(x) = \frac{0 \cdot (1+e^{-x}) - 2(-e^{-x})}{(1+e^{-x})^2} = \frac{2e^{-x}}{(1+e^{-x})^2}. \]

Step 2: Determine Increasing/Decreasing Intervals

The function \( f(x) \) is increasing where \( f'(x) > 0 \) and decreasing where \( f'(x) < 0 \). Since \( f'(x) = \frac{2e^{-x}}{(1+e^{-x})^2} \) and \( e^{-x} > 0 \) for all \( x \), \( f'(x) > 0 \) for all \( x \). Therefore, \( f(x) \) is increasing for all real numbers, \( x \in \mathbf{R} \).

Step 3: Find the Second Derivative

Next, we find the second derivative \( f''(x) \) to determine concavity. Using the quotient rule again on \( f'(x) = \frac{2e^{-x}}{(1+e^{-x})^2} \): Let \( u = 2e^{-x} \) and \( v = (1+e^{-x})^2 \). Then \( u' = -2e^{-x} \) and \( v' = -2(1+e^{-x})(e^{-x}) \). Using these, we have:\[ f''(x) = \frac{(-2e^{-x})(1+e^{-x})^2 - 2e^{-x}(-2(1+e^{-x})e^{-x})}{(1+e^{-x})^4}. \] Simplify to get the second derivative:\[ f''(x) = \frac{2e^{-x}(e^{-x} - 1)}{(1+e^{-x})^3}. \]

Step 4: Determine Concavity and Inflection Points

Function \( f(x) \) is concave up where \( f''(x) > 0 \) and concave down where \( f''(x) < 0 \). Solving \( 2e^{-x}(e^{-x} - 1) = 0 \) gives \( e^{-x} - 1 = 0 \), or \( x = 0 \). Test points around \( x = 0 \): \( f''(x) \) changes from negative to positive, showing an inflection point at \( x = 0 \). Thus, \( f(x) \) is concave down on \( (-\infty, 0) \) and concave up on \( (0, \infty) \).

Step 5: Evaluate Limits at Infinity

To find horizontal asymptotes, compute limits at infinity: \( \lim _{x \to \infty} f(x) = \lim _{x \to \infty} \frac{2}{1+e^{-x}} = \frac{2}{1+0} = 2 \). So, there is a horizontal asymptote at \( y = 2 \).

Step 6: Evaluate Limits at Negative Infinity

Find the limit as \( x \to -\infty \): \( \lim _{x \to -\infty} f(x) = \lim _{x \to -\infty} \frac{2}{1+e^{-x}} = \frac{2}{1+\infty} = 0 \). Therefore, there is a horizontal asymptote at \( y = 0 \).

Step 7: Sketch the Graph

The function \( f(x) = \frac{2}{1+e^{-x}} \) is increasing across its entire domain with horizontal asymptotes at \( y = 2 \) for \( x \to \infty \) and \( y = 0 \) for \( x \to -\infty \). It has an inflection point at \( x = 0 \), where it changes from concave down to concave up. Plot these features to accurately sketch the graph.

Key Concepts

Increasing and Decreasing Functions

Understanding whether a function is increasing or decreasing is key to analyzing its behavior. For the function given, \( f(x) = \frac{2}{1+e^{-x}} \), this is determined by examining the first derivative: \( f'(x) = \frac{2e^{-x}}{(1+e^{-x})^2} \).
The derivative tells us how the function behaves as \( x \) changes. Here are some important points to help understand this:

• The function is increasing where \( f'(x) > 0 \).
• The exponential term, \( e^{-x} \), is always positive for all \( x \).
• The denominator \((1+e^{-x})^2\) is also always positive because a square is always non-negative.

Putting this all together, since both the numerator and denominator of \( f'(x) \) are positive for all \( x \), \( f(x) \) is increasing for all real numbers. Understanding these critical points ensures a solid foundation in calculus analysis.

Concavity and Inflection Points

Concavity indicates how a function curves, which can be upwards or downwards. We explore this using the second derivative, \( f''(x) \). For \( f(x) = \frac{2}{1+e^{-x}} \), the second derivative is calculated as \[ f''(x) = \frac{2e^{-x}(e^{-x} - 1)}{(1+e^{-x})^3} \].
Concavity depends on the sign of \( f''(x) \):

• When \( f''(x) > 0 \), the function is concave up.
• When \( f''(x) < 0 \), the function is concave down.

Solving \( f''(x) = 0 \), we find \( x = 0 \) as a potential point of inflection. To confirm, test intervals around \( x = 0 \):

• For \( x < 0 \), \( f''(x) < 0 \); the function is concave down.
• For \( x > 0 \), \( f''(x) > 0 \); the function is concave up.

This change in concavity at \( x = 0 \) confirms it as an inflection point, where \( f(x) \) changes curvature.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a function as \( x \) approaches infinity or negative infinity. For \( f(x) = \frac{2}{1+e^{-x}} \), we find these limits to understand its long-term behavior:

• As \( x \to \infty \): The term \( e^{-x} \to 0 \) because exponentials drop towards zero.
• This simplifies to \( f(x) \approx \frac{2}{1+0} = 2 \). Thus, \( y = 2 \) is a horizontal asymptote.
• As \( x \to -\infty \): The term \( e^{-x} \to \infty \).
• Thus, \( f(x) \approx \frac{2}{\infty} = 0 \). Therefore, \( y = 0 \) is another horizontal asymptote.

Recognizing these horizontal asymptotes helps in visualizing how \( f(x) \) behaves at the far ends of the graph, providing a clear picture of its trend over the entire real line.