Problem 27 - Calculus Early Transcendentals

Question

Find: (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. \(f(x)=e^{-x^{2} / 2}\)

Step-by-Step Solution

Answer

Increasing on \((-\infty, 0)\), decreasing on \((0, \infty)\), concave up on \((-\infty, -1) \cup (0, 1)\), concave down on \((-1, 0) \cup (1, \infty)\). Inflection points at \(x = -1\) and \(x = 1\).

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, first find the derivative of the function. For \( f(x)=e^{-x^{2}/2} \), apply the chain rule. The derivative is \( f'(x) = -x \cdot e^{-x^{2}/2} \).

Step 2: Determine Critical Points

Set the first derivative equal to zero to find critical points. Solve \( -x \cdot e^{-x^{2}/2} = 0 \). Since \( e^{-x^{2}/2} \) is never zero, solve \( -x = 0 \), yielding \( x = 0 \).

Step 3: Identify Increasing and Decreasing Intervals

Use the sign of \( f'(x)\) to determine intervals of increase and decrease. For \( x < 0 \), \( f'(x) > 0 \) (since \( -x \) is positive), so the function is increasing. For \( x > 0 \), \( f'(x) < 0 \) (since \( -x \) is negative), so the function is decreasing. The function is increasing on \((-fty, 0)\) and decreasing on \((0, fty)\).

Step 4: Find the Second Derivative

To determine concavity, find the second derivative. Differentiate \( f'(x) = -x \cdot e^{-x^{2}/2} \) again using the product rule to get \( f''(x) = (x^2 -1)e^{-x^{2}/2} \).

Step 5: Analyze Concavity

Set \( f''(x) \) to zero to find potential inflection points: \( (x^2 - 1)e^{-x^{2}/2} = 0 \). This occurs when \( x^2 - 1 = 0 \), so \( x = \pm 1 \). To determine concavity, consider intervals around these points. For \( x < -1 \) and \( 0 < x < 1 \), \( f''(x) > 0 \); and for \( -1 < x < 0 \) and \( x > 1 \), \( f''(x) < 0 \). Thus, concave up on \((-fty, -1) \cup (0, 1)\) and concave down on \((-1, 0) \cup (1, fty)\).

Step 6: Identify Inflection Points

Inflection points occur where the second derivative changes sign. Here, changes at \( x = -1, 1 \). Evaluate the function at these points: \( f(-1) = e^{-1/2} \) and \( f(1) = e^{-1/2} \). So, inflection points are at \((-1, e^{-1/2})\) and \((1, e^{-1/2})\).

Key Concepts

Derivative

Understanding derivatives is vital in calculus, as they represent how a function changes at any given point. Derivatives describe the rate of change or slope of the function's curve. For the function \(f(x) = e^{-x^2/2}\), we use the chain rule to find the derivative. The chain rule helps us differentiate composite functions. The first derivative of this function is \(f'(x) = -x \cdot e^{-x^2/2}\). This expression tells us the instantaneous rate of change of \(f(x)\) regarding \(x\).

• The derivative can inform us about the behavior of a graph such as where it increases or decreases.
• Positive values of the derivative indicate an increasing function.
• Negative values mean the function is decreasing.

Critical Points

Critical points occur where the first derivative is zero or undefined. These points are essential because they can indicate where a function changes from increasing to decreasing or vice versa. In practice, we set \(f'(x) = -x \cdot e^{-x^2/2} = 0\). Since exponential functions like \(e^{-x^2/2}\) do not equal zero, we solve for \(x\) where \(-x = 0\). Thus, the critical point is at \(x = 0\).

• At \(x = 0\), the behavior of the function shifts.
• These points are potential minimums or maximums on the curve.

Understanding critical points allows us to analyze the function's intervals of increase and decrease.

Concave Up

A function is said to be concave up on an interval if its second derivative is positive over that interval. When a graph is concave up, it looks like the open side of a bowl. To identify these intervals for \(f(x) = e^{-x^2/2}\), we need the second derivative: \(f''(x) = (x^2 - 1)e^{-x^2/2}\). By setting \(f''(x) > 0\), we find that it is concave up for \(x < -1\) and \(0 < x < 1\).

• When \(f''(x) > 0\), the graph holds a shape bending upwards.
• This indicates that the slope of \(f'(x)\) is increasing.

Knowing where a function is concave up informs us about its curvature.

Concave Down

Conversely, concave down intervals occur when the second derivative is negative over a specified interval. These parts of the graph appear like an upside-down bowl. For the function \(f(x) = e^{-x^2/2}\), it is concave down when \(f''(x) = (x^2 - 1)e^{-x^2/2} < 0\). This holds for the intervals \(-1 < x < 0\) and \(x > 1\).

• In concave down sections, the slope of the function's derivative decreases.
• The graph appears to bend downwards, influencing the shape significantly.

Identifying where a function is concave down is crucial for understanding its behavior and geometry.

Inflection Points

Inflection points are fascinating because they mark where the function changes concavity. This occurs where the second derivative changes signs. For \(f(x) = e^{-x^2/2}\), inflection points occur at \(x = -1\) and \(x = 1\). Evaluating \(f(x)\) at these points gives the coordinates \((-1, e^{-1/2})\) and \((1, e^{-1/2})\).

• Inflection points signify where the graph transitions from concave up to concave down or vice versa.
• The sign change in \(f''(x)\) indicates these turning points.

Detecting inflection points helps in sketching the graph accurately and comprehensively understanding the function's overall shape.