Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 19-28, 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 graph increases on \((-\infty, 0)\), decreases on \((0, \infty)\), is concave up on \((-\infty, -\sqrt{1/2}) \cup (\sqrt{1/2}, \infty)\), and concave down on \((-\sqrt{1/2}, \sqrt{1/2})\).

Step 1: Find the First Derivative

First, to determine where the function is increasing and decreasing, we need the first derivative. The function is \( f(x) = e^{-x^2} \). Using the chain rule, the first derivative is \( f'(x) = -2xe^{-x^2} \).

Step 2: Find Critical Points

Set the first derivative \( f'(x) = -2xe^{-x^2} \) to zero and solve for \( x \). This gives \( -2xe^{-x^2} = 0 \), which simplifies to \( x = 0 \) as the only critical point.

Step 3: Determine Intervals for Increasing/Decreasing

Use a sign chart for \( f'(x) \). For \( x < 0 \), \( f'(x) = -2xe^{-x^2} \) is positive, so the function is increasing. For \( x > 0 \), \( f'(x) \) is negative, so the function is decreasing.

Step 4: Find the Second Derivative

To determine concavity, find the second derivative. Differentiate \( f'(x) = -2xe^{-x^2} \) using the product rule to get \( f''(x) = (4x^2 - 2)e^{-x^2} \).

Step 5: Find Inflection Points

Set the second derivative \( f''(x) = (4x^2 - 2)e^{-x^2} \) to zero to find inflection points. This simplifies to \( 4x^2 - 2 = 0 \), giving \( x = \pm \sqrt{\frac{1}{2}} \).

Step 6: Determine Intervals for Concavity

Use a sign chart for \( f''(x) \). For \( x < -\sqrt{\frac{1}{2}} \), \( f''(x) > 0 \), so the function is concave up. Between \( -\sqrt{\frac{1}{2}} \) and \( \sqrt{\frac{1}{2}} \), \( f''(x) < 0 \), so it is concave down. For \( x > \sqrt{\frac{1}{2}} \), \( f''(x) > 0 \), hence the function is concave up again.

Step 7: Sketch the Graph

Using all information: The function is increasing for \( x < 0 \), decreasing for \( x > 0 \), concave up for \( x < -\sqrt{\frac{1}{2}} \) and \( x > \sqrt{\frac{1}{2}} \), and concave down in between. Sketch the corresponding graph with a peak at \( x = 0 \).

Key Concepts

First derivative

To analyze a function's behavior, determine the first derivative. It gives us the rate at which the function changes. For the function \( f(x) = e^{-x^2} \), applying the chain rule, we find the first derivative: \( f'(x) = -2xe^{-x^2} \).

• Purpose: The first derivative helps identify intervals where the function increases or decreases.
• Interpreting Signs:
  • If \( f'(x) > 0 \), the function is increasing.
  • If \( f'(x) < 0 \), the function is decreasing.

Looking at the computed \( f'(x) \), it's positive when \( x < 0 \) and negative when \( x > 0 \), meaning the function rises on the left and falls on the right side of the point \( x = 0 \). This creates a peak at \( x = 0 \). By understanding this derivative, we can visually interpret how the function trends over different intervals.

Second derivative

The second derivative provides information on a function’s concavity, which indicates how the curve bends. For the given function \( f(x) = e^{-x^2} \), the second derivative is found by differentiating the first derivative. Applying the product rule yields \( f''(x) = (4x^2 - 2)e^{-x^2} \).

• Purpose: The second derivative tells us which parts of the function's graph are concave up or down.
• Interpreting Signs:
  • If \( f''(x) > 0 \), the graph is concave up, often resembling a cup.
  • If \( f''(x) < 0 \), the graph is concave down, similar to a dome.

The second derivative shows that sections where \( 4x^2 > 2 \) are concave up, while other areas are concave down. This curvature information is crucial in creating accurate graph depictions of functions.

Critical points

Critical points are vital in analyzing functions as they inform us about any peaks, valleys, or flat spots. These points occur where the first derivative is zero or undefined. For \( f(x) = e^{-x^2} \), setting \( f'(x) = -2xe^{-x^2} \) to zero gives the sole critical point at \( x = 0 \).

• Purpose: Identify potential local maximums or minimums.
• Steps to Identify:
  • Solve \( f'(x) = 0 \) for \( x \).
  • Check where \( f'(x) \) is undefined, if applicable.

In this problem, since the critical point is found at \( x = 0 \), the function has a peak there before changing from increasing to decreasing. This turning point greatly affects the graph shape.

Concavity

Concavity is about the way a graph bends and can be analyzed using the second derivative. For \( f(x) = e^{-x^2} \), our second derivative is \( f''(x) = (4x^2 - 2)e^{-x^2} \).

• Concave Up: The graph bends upwards.
• Concave Down: The graph bends downwards.

Finding intervals for concavity involves:

• Setting \( f''(x) = 0 \) and solving for \( x \).
• Creating a sign chart to analyze sign changes in \( f''(x) \).

In our specific case:- Graph is concave up outside the range \( x = -\sqrt{\frac{1}{2}} \) to \( x = \sqrt{\frac{1}{2}} \).- It's concave down between these points.Knowing these intervals aids in sketching a more accurate graph and predicting point behavior.

Inflection points

Inflection points are where the graph transitions concavity – switching from concave up to concave down or vice versa. Detect these points by setting the second derivative equal to zero. For \( f(x) = e^{-x^2} \), this occurs when \( f''(x) = (4x^2 - 2)e^{-x^2} = 0 \), leading to solutions \( x = \pm \sqrt{\frac{1}{2}} \).

• Purpose: Inflection points mark changes in the graph’s bending direction.
• Steps to Find:
  • Solve \( f''(x) = 0 \) for \( x \).

Beyond marking where concavity changes, these points help flesh out graph details and are key in accurately interpreting function shapes. Studying them enhances our visual and conceptual understanding of the function's overall behavior.