Problem 13 - Calculus Single Variable

Question

Use the first derivative to determine the intervals on which the given function \(f\) is increasing and on which \(f\) is decreasing. At each point \(c\) with \(f^{\prime}(c)=0,\) use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=\left(x^{2}-3\right) /(x+2) $$

Step-by-Step Solution

Answer

\( f(x) \) is increasing on \((-\infty, -3)\), \((-3, -2)\), and \((-1, \infty)\), decreasing on \((-2, -1)\), with a local maximum at \(x = -3\) and a local minimum at \(x = -1\).

Step 1: Compute the First Derivative

To find where the function is increasing or decreasing, we first compute the first derivative of \( f(x) = \frac{x^2 - 3}{x + 2} \). We use the quotient rule: if \( u = x^2 - 3 \) and \( v = x + 2 \), then \( f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \).Compute the derivatives: \( u' = 2x \) and \( v' = 1 \).Then, \( f'(x) = \frac{(x + 2)(2x) - (x^2 - 3)(1)}{(x + 2)^2} = \frac{2x^2 + 4x - x^2 + 3}{(x + 2)^2} = \frac{x^2 + 4x + 3}{(x + 2)^2} \).

Step 2: Find Critical Points

The critical points occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined. Set \( f'(x) = 0 \):\[ x^2 + 4x + 3 = 0 \]Factor it: \( (x + 1)(x + 3) = 0 \), so \( x = -1 \) and \( x = -3 \) are critical points.Since \((x + 2)^2\) is in the denominator of \( f'(x) \), \( f'(x) \) is undefined at \( x = -2 \).

Step 3: Determine the Intervals of Increase/Decrease

Test the intervals determined by the critical points and the point where the derivative is undefined: \((-\infty, -3)\), \((-3, -2)\), \((-2, -1)\), and \((-1, \infty)\).1. Choose a test point in each interval: use \( x = -4 \), \( x = -2.5 \), \( x = -1.5 \), and \( x = 0 \).2. Calculate the sign of \( f'(x) \): - For \( x = -4 \), \( f'(-4) = \frac{16 - 16 + 3}{4} = \frac{3}{4} > 0 \) (increasing). - For \( x = -2.5 \), \( f'(-2.5) = \frac{6.25 \times 1.5 + 3}{(0.5)^2} = 8.5 \) (increasing). - For \( x = -1.5 \), \( f'(-1.5) = \frac{2.25 \times 2.5 + 3}{0.25} = -22.5 \) (decreasing). - For \( x = 0 \), \( f'(0) = \frac{3}{4} > 0 \) (increasing).Thus, \( f(x) \) is increasing on \( (-\infty, -3) \), \((-3, -2)\), and \((-1, \infty) \), and decreasing on \( (-2, -1) \).

Step 4: Apply the First Derivative Test

At \( x = -3 \), \( f'(x) \) changes from positive to negative, indicating a local maximum.At \( x = -1 \), \( f'(x) \) changes from negative to positive, indicating a local minimum.The point \( x = -2 \) is not in the domain of \( f(x) \) due to division by zero and is not considered for local extrema.

Key Concepts

First Derivative

The first derivative of a function provides essential information about the function's behavior. Specifically, it describes how the function's value changes as the input changes. When we calculate the first derivative of a function, we essentially find a new function that expresses the rate of change of the original function. This derivative is symbolically represented as \( f'(x) \).
The process involves using rules of differentiation, like the power rule, product rule, or quotient rule, depending on the form of the function. In this exercise, we use the quotient rule because the function \( f(x) = \frac{x^2 - 3}{x + 2} \) is a quotient. The quotient rule is applied when you have one function divided by another, given by:

• If \( u = x^2 - 3 \) and \( v = x + 2 \), then \( f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \).

Here, \( u' = 2x \) and \( v' = 1 \), leading us to find that \( f'(x) = \frac{(x + 2)(2x) - (x^2 - 3)(1)}{(x + 2)^2} \). This is simplified to provide insight into the nature of the function.

Critical Points

Critical points of a function occur where its first derivative equals zero or is undefined. These points provide information on the potential maxima, minima, or points of inflection for the function.
For this exercise, to find critical points, we set \( f'(x) = 0 \). Solving for zero, we use:

• \( x^2 + 4x + 3 = 0 \)

This is a quadratic equation that can be factored as \((x + 1)(x + 3) = 0 \). Thus, the critical points are at \( x = -1 \) and \( x = -3 \).
Additionally, since our original function is a rational function, the derivative is undefined when the denominator equals zero. In this case, \( f'(x) \) is undefined at \( x = -2 \) because it would require division by zero, indicating a potential vertical asymptote or discontinuity which cannot host a local extremum.

Increasing and Decreasing Intervals

To determine increasing and decreasing intervals of a function, we look at the sign of its first derivative, \( f'(x) \). When \( f'(x) > 0 \), the function is increasing, and when \( f'(x) < 0 \), it is decreasing.
From the critical points and where the derivative is undefined, we set test intervals:

• \((-, -3)\), \((-3, -2)\), \((-2, -1)\), and \((-1, )\)

We choose test points in each interval:

• For \(x = -4\), \( f'(-4) > 0 \), indicating an increasing interval.
• For \(x = -2.5\), \( f'(-2.5) > 0 \), still increasing.
• For \(x = -1.5\), \( f'(-1.5) < 0 \), indicating a decreasing interval.
• For \(x = 0\), \( f'(0) > 0 \), indicates another increasing interval.

Hence, the function increases on \((-, -3)\), \((-3, -2)\), and \((-1, )\), and decreases on \((-2, -1)\).

First Derivative Test

The First Derivative Test helps us determine if each critical point is a local extremum:

• Local Maximum: At a critical point, if \( f'(x) \) changes from positive to negative, the function has a local maximum there.
• Local Minimum: At a critical point, if \( f'(x) \) changes from negative to positive, the function has a local minimum.

Applying this to our critical points:

• At \( x = -3 \), \( f'(x) \) shifts from positive to negative, indicating a local maximum at \( x = -3 \).
• At \( x = -1 \), \( f'(x) \) changes from negative to positive, indicating a local minimum at \( x = -1 \).

Note: The point where the derivative is undefined, \( x = -2 \), is not considered for local extrema, as it falls outside the domain of the function due to division by zero.