Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine where the graph of the given function is increasing, decreasing, concave up, and concave down. Then sketch the graph (see Example 4). $$ f(x)=e^{-x^{2}} $$

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, 0)\), decreasing on \((0, \infty)\); concave up on \((-\infty, -\frac{1}{\sqrt{2}})\) and \((\frac{1}{\sqrt{2}}, \infty)\), concave down on \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\).

Step 1: Find the first derivative

To determine where the function is increasing or decreasing, we need to find the first derivative of the function, \( f(x) = e^{-x^2} \). Use the chain rule to differentiate:\[ f'(x) = -2xe^{-x^2} \].

Step 2: Identify critical points

Critical points occur where the first derivative is zero or undefined. Solve the equation:\[ -2xe^{-x^2} = 0 \]Since \( e^{-x^2} eq 0 \) for all \( x \), we set \( -2x = 0 \), giving \( x = 0 \).

Step 3: Determine intervals of increase and decrease

Assess the sign of \( f'(x) \) on either side of the critical point \( x = 0 \):- For \( x < 0 \), \( f'(x) = -2xe^{-x^2} > 0 \) (since \( x \) and \( -2x \) are positive).- For \( x > 0 \), \( f'(x) = -2xe^{-x^2} < 0 \) (since \( x \) and \( -2x \) are negative).Thus, the function is increasing on \( (-\infty, 0) \) and decreasing on \( (0, \infty) \).

Step 4: Find the second derivative

The concavity of the function is determined by the second derivative. Differentiate \( f'(x) = -2xe^{-x^2} \) to get the second derivative. Using the product and chain rule:\[ f''(x) = (4x^2 - 2)e^{-x^2} \].

Step 5: Identify inflection points

Solve \( f''(x) = 0 \) for points of possible inflection:\[ (4x^2 - 2)e^{-x^2} = 0 \]This simplifies to \( 4x^2 - 2 = 0 \), giving \( x^2 = \frac{1}{2} \) or \( x = \pm \frac{1}{\sqrt{2}} \).

Step 6: Determine intervals of concavity

Test the concavity by checking the sign of \( f''(x) \) around the inflection points \( x = \pm \frac{1}{\sqrt{2}} \):- For \( x < -\frac{1}{\sqrt{2}} \), \( f''(x) > 0 \), the function is concave up.- For \( -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \), \( f''(x) < 0 \), the function is concave down.- For \( x > \frac{1}{\sqrt{2}} \), \( f''(x) > 0 \), the function is concave up.

Key Concepts

Derivative

The derivative of a function gives us the rate at which the function’s value changes with respect to changes in its input value. For the function \( f(x) = e^{-x^2} \), we need to find the first derivative to understand the behavior of the function, particularly when it increases or decreases. The first derivative is calculated using the chain rule. Here, the first derivative is \( f'(x) = -2xe^{-x^2} \). This expression tells us how the slope of the tangent to the function's graph changes as \( x \) changes. The sign of \( f'(x) \) plays a crucial role in determining the intervals where the function is increasing or decreasing. A positive \( f'(x) \) indicates that the function is increasing, while a negative \( f'(x) \) suggests the function is decreasing.

Concavity

Concavity refers to the direction in which a function curves. To determine concavity, we need to look at the second derivative of the function. The second derivative \( f''(x) \) tells us how the rate of change of the slope, or the derivative, itself changes. In the given problem, the second derivative \( f''(x) = (4x^2 - 2)e^{-x^2} \) is found using both the product and chain rules. If \( f''(x) > 0 \), the function is said to be concave up, resembling a U-shape. Conversely, if \( f''(x) < 0 \), the function is concave down, similar to an upside-down U-shape. Understanding concavity helps in sketching the graph and analyzing the function's behavior around certain points.

Increasing Decreasing Intervals

Once we have the derivative, we can determine where the function is increasing or decreasing. These intervals are identified by analyzing the sign of the first derivative \( f'(x) \). In this case, solve \( -2xe^{-x^2} = 0 \) to find critical points which are \( x = 0 \) in this scenario. By examining the sign of \( f'(x) \) in different intervals around \( x = 0 \), we observe:

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

Thus, the function increases on \( (-\infty, 0) \) and decreases on \( (0, \infty) \). This behavior aids in plotting the function's broad shape.

Critical Points

Critical points of a function are values of \( x \) where the first derivative is zero or undefined. These points are significant because they can indicate local maxima, minima, or points of inflection. For the function \( f(x) = e^{-x^2} \), the first derivative is zero at \( x = 0 \). To classify this critical point, assess the sign of \( f'(x) \) on intervals around \( x = 0 \):

• Before \( x = 0 \), \( f'(x) > 0 \) implies the function is rising.
• After \( x = 0 \), \( f'(x) < 0 \) implies the function is falling.

This indicates that \( x = 0 \) is a local maximum. Critical points are essential in understanding the peaks and troughs of a function's graph.

Inflection Points

Inflection points are where the concavity of a function changes, from concave up to concave down, or vice versa. To find these points, set the second derivative to zero and solve for \( x \). For \( f(x) = e^{-x^2} \), solving \( (4x^2 - 2)e^{-x^2} = 0 \) gives \( x^2 = \frac{1}{2} \), thus \( x = \pm \frac{1}{\sqrt{2}} \). At these inflection points:

• The function changes from concave up to concave down between \( x = -\frac{1}{\sqrt{2}} \) and \( \frac{1}{\sqrt{2}} \).
• It then returns to concave up for \( x > \frac{1}{\sqrt{2}} \).

Identifying inflection points helps clarify the subtle bends and curves of the graph, offering a complete picture of the function's structure.