Problem 29 - 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. \(f(x)=\ln \sqrt{x^{2}+4}\)

Step-by-Step Solution

Answer

(a) Increasing on \((0, \infty)\); (b) Decreasing on \((-\infty, 0)\); (c) Concave up on \((-2, 2)\); (d) Concave down on \((-\infty, -2) \cup (2, \infty)\); (e) Inflection points at \((-2, \ln \sqrt{8})\) and \((2, \ln \sqrt{8})\).

Step 1: Find the First Derivative

To find where the function is increasing or decreasing, compute the first derivative of the function. Start by simplifying the function: \( f(x) = \ln(x^{2} + 4)^{1/2} \), which can be rewritten as \( f(x) = \frac{1}{2} \ln(x^{2} + 4) \). Now differentiate: \( f'(x) = \frac{1}{2} \cdot \frac{1}{x^{2} + 4} \cdot 2x = \frac{x}{x^{2} + 4} \).

Step 2: Determine Intervals of Increase and Decrease

Set the first derivative equal to zero to find critical points: \( \frac{x}{x^{2} + 4} = 0 \). Solving this, we find \( x = 0 \). Analyze the sign of \( f'(x) \) around this point. For \( x < 0 \), \( f'(x) < 0 \), and for \( x > 0 \), \( f'(x) > 0 \). Thus, the function is decreasing on \( (-\infty, 0) \) and increasing on \( (0, \infty) \).

Step 3: Find the Second Derivative

To determine concavity, calculate the second derivative. Differentiate \( f'(x) = \frac{x}{x^{2} + 4} \) using the quotient rule: \[f''(x) = \frac{(x^{2} + 4) \cdot 1 - x \cdot 2x}{(x^{2} + 4)^{2}} = \frac{x^{2} + 4 - 2x^{2}}{(x^{2} + 4)^{2}} = \frac{-x^{2} + 4}{(x^{2} + 4)^{2}}.\]

Step 4: Determine Intervals of Concavity

Set the second derivative equal to zero to find possible inflection points: \( -x^{2} + 4 = 0 \) gives \( x^{2} = 4 \), so solutions are \( x = -2 \) and \( x = 2 \). Test intervals around these points. For \( x < -2 \) and \( x > 2 \), \( f''(x) < 0 \) (concave down), and for \( -2 < x < 2 \), \( f''(x) > 0 \) (concave up). Thus, the function is concave up on \((-2, 2)\) and concave down on \((-\infty, -2) \cup (2, \infty)\).

Step 5: Identify Inflection Points

Inflection points occur where the concavity changes, which happens at \( x = -2 \) and \( x = 2 \). Evaluate \( f(x) \) at these x-values for the inflection points: \( f(-2) = \ln \sqrt{(-2)^{2}+4} = \ln \sqrt{8} \) and similarly, \( f(2) = \ln \sqrt{8} \). Therefore, the inflection points are \((-2, \ln \sqrt{8})\) and \((2, \ln \sqrt{8})\).

Key Concepts

Derivatives

Derivatives are a core concept in calculus that help us determine the rate at which a function changes. For a given function, the derivative, often denoted as \( f'(x) \), gives us the slope of the tangent line to the curve at any point \( x \). This is incredibly useful for understanding how a function behaves.

By differentiating our function \( f(x) = \ln \sqrt{x^2+4} \), we first express it as \( \frac{1}{2} \ln(x^2 + 4) \). Differentiating using the chain rule, we find \( f'(x) = \frac{x}{x^2 + 4} \).

Understanding how the first derivative works helps us discern when a function is increasing or decreasing. In essence, \( f'(x) > 0 \) when the function is increasing, and \( f'(x) < 0 \) when it is decreasing.

Function Behavior

The behavior of a function, such as where it is increasing or decreasing, is critical in understanding its nature. By analyzing the first derivative, \( f'(x) \), we can find critical points where the function transitions from increasing to decreasing or vice versa.

For our function, by setting \( f'(x) = \frac{x}{x^2 + 4} = 0 \), we find that the critical point is at \( x = 0 \). Testing values around this critical point, we observe:

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

Consequently, this function is decreasing on the interval \((-\infty, 0)\) and increasing on \((0, \infty)\). Understanding this behavior is essential for graphing and analyzing real-world scenarios where such patterns occur.

Concavity

Concavity refers to the curvature direction of a function. A function is concave up when it curves upwards like a cup and concave down when it curves downwards. To determine concavity, we use the second derivative, \( f''(x) \).

For our function \( f'(x) = \frac{x}{x^2 + 4} \), the second derivative is \( f''(x) = \frac{-x^2 + 4}{(x^2 + 4)^2} \).

By testing intervals around the solutions to \( f''(x) = 0 \), identified as \( x = -2 \) and \( x = 2 \), we find:

• The function is concave up on \((-2, 2)\), where \( f''(x) > 0 \).
• It is concave down on \((-fty, -2)\) and \((2, fty)\), where \( f''(x) < 0 \).

This analysis helps identify the function's curvature intervals and adjusts modeling predictions accordingly.

Inflection Points

Inflection points are where a function changes its concavity; from concave up to concave down or vice versa. These points are significant because they often represent transitions in the behavior of real-world phenomena.

To find these, we set the second derivative equal to zero and solve: \( -x^2 + 4 = 0 \), resulting in \( x = -2 \) and \( x = 2 \). These are the x-coordinates of the inflection points.

Evaluating the original function at these points, we calculate:

• At \( x = -2 \), \( f(-2) = \ln \sqrt{8} \).
• At \( x = 2 \), \( f(2) = \ln \sqrt{8} \).

Thus, the inflection points are \(-2, \ln \sqrt{8}\) and \(2, \ln \sqrt{8}\). Recognizing these points is crucial in understanding and predicting switching behaviors in a function.