Problem 59 - Calculus and its applications

Question

Graph each function. Then determine any critical values, inflection points, intervals over which the function is increasing or decreasing, and the concavity. $$ f(x)=\frac{1}{2} e^{-x} $$

Step-by-Step Solution

Answer

The function is always decreasing and concave up, with no critical or inflection points.

Step 1: Analyze the Function

The given function is \( f(x) = \frac{1}{2} e^{-x} \). This is an exponential decay function due to the negative exponent on \( e \). It's useful to begin by understanding its behavior: as \( x \) increases, \( f(x) \) approaches 0. As \( x \) decreases, \( f(x) \) increases towards \( \frac{1}{2} \).

Step 2: Find the Derivative

To explore where the function is increasing or decreasing, find the first derivative: \[ f'(x) = \frac{d}{dx} \left( \frac{1}{2} e^{-x} \right) = -\frac{1}{2} e^{-x}. \] This derivative is always negative, meaning the function is always decreasing.

Step 3: Determine Critical Values

Critical values occur where \( f'(x) = 0 \) or is undefined. Since \( -\frac{1}{2} e^{-x} eq 0 \) and is never undefined, there are no critical points because the derivative is always negative.

Step 4: Check for Concavity

To find concavity, compute the second derivative: \[ f''(x) = \frac{d}{dx} \left( -\frac{1}{2} e^{-x} \right) = \frac{1}{2} e^{-x}. \] Since the second derivative is positive for all \( x \), the function is concave up everywhere.

Step 5: Identify Inflection Points

An inflection point occurs where the concavity changes, which happens if \( f''(x) = 0 \). In this function, \( \frac{1}{2} e^{-x} eq 0 \) for any \( x \), so there are no inflection points.

Key Concepts

Exponential Functions

Exponential functions are a key concept in calculus and mathematics in general. These functions are defined by the formula \( f(x) = a e^{bx} \), where \( a \) and \( b \) are constants, and \( e \) is the mathematical constant approximately equal to 2.71828. In any exponential function, the base \( e \) is raised to a power, which is often a variable multiplied by a coefficient.

Exponential functions can represent growth or decay. If the exponent is positive, the function grows as \( x \) increases. Conversely, if the exponent is negative, we observe exponential decay, as is the case with our function \( f(x) = \frac{1}{2} e^{-x} \). This negative exponent suggests a decay, meaning the function value decreases as \( x \) increases. Such behavior is critical in modeling natural phenomena like radioactive decay or cooling processes.

• Growth: \( b > 0 \)
• Decay: \( b < 0 \)

This decay nature helps predict that the value of \( f(x) \) asymptotically approaches zero as \( x \) heads towards infinity.

Derivative

The derivative of a function provides us with critical insights into its behavior. It represents the rate of change of a function concerning its variable. For the function \( f(x) = \frac{1}{2} e^{-x} \), the derivative \( f'(x) \) is calculated using the power rule combined with the derivative of the exponential function.

For this problem, the first derivative is \( f'(x) = -\frac{1}{2} e^{-x} \). The negative sign indicates that the function is always decreasing. A decreasing function means that as the input \( x \) increases, the output \( f(x) \) decreases. In our function, no matter the value of \( x \), the first derivative remains negative.

• The derivative is negative: the function is decreasing.
• Critical values: \( f'(x) = 0 \), not applicable here.

The derivative is one of the key tools in calculus for understanding whether a function is increasing or decreasing on specific intervals.

Concavity

Concavity describes how a function curves. This is determined by the second derivative of the function. If \( f''(x) > 0 \), the function is concave up, resembling a smiling curve. If \( f''(x) < 0 \), it's concave down, resembling a frowning curve. For this function, \( f''(x) = \frac{1}{2} e^{-x} \), which is always positive due to the nature of the exponential function.

Because \( f''(x) \) is positive for all \( x \), our function \( f(x) = \frac{1}{2} e^{-x} \) is concave up throughout its domain. This consistent concavity leads to the following interpretations:

• The function continuously curves upwards.
• Second derivative helps indicate the shape of the graph.

This concave-up nature might suggest that the value of the function approaching its horizontal asymptote can do so gently, without any abrupt changes in direction.

Inflection Points

Inflection points are points on a graph where the concavity changes. They are found by setting the second derivative equal to zero and solving for \( x \). An inflection point indicates a transition from concave up to concave down, or vice versa.

For our function \( f(x) = \frac{1}{2} e^{-x} \), we have established that \( f''(x) = \frac{1}{2} e^{-x} > 0 \) for all \( x \). This means the concavity is constant (concave up) and does not change at any point on the function. Thus, there are no inflection points.

• No change in concavity implies no inflection points.
• Always check \( f''(x) = 0 \) to identify potential inflection points.

Absence of inflection points usually indicates a smooth, consistent curve in the graph of the function.