Problem 16 - Calculus - AP Edition

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. $$f(x)=5-4 x-x^{2}$$

Step-by-Step Solution

Answer

(a) Increasing on \((- fty, -2)\), (b) Decreasing on \((-2, fty)\), (c) None, (d) Concave down on \((- fty, fty)\), (e) No inflection points.

Step 1: Find the First Derivative

To determine the intervals where the function is increasing or decreasing, we start by taking the first derivative of the function. Given function: \[ f(x) = 5 - 4x - x^2 \]First derivative of the function: \[ f'(x) = \frac{d}{dx}(5 - 4x - x^2) = -4 - 2x \]

Step 2: Find Critical Points

To find the critical points, set the first derivative equal to zero and solve for \(x\). These points will help determine the intervals where \(f\) is increasing or decreasing.\[ -4 - 2x = 0 \]Solving for \(x\), we have:\[ 2x = -4 \]\[ x = -2 \]

Step 3: Test Intervals Based on Critical Points

Pick test points in the intervals defined by the critical points to determine where \(f\) is increasing or decreasing.Intervals: \((-\infty, -2)\) and \((-2, \infty)\).Test point for \((-\infty, -2)\) is \(x = -3\):\[ f'(-3) = -4 - 2(-3) = 2 \] (positive, so \(f\) is increasing)Test point for \((-2, \infty)\) is \(x = 0\):\[ f'(0) = -4 - 2(0) = -4 \] (negative, so \(f\) is decreasing)

Step 4: Find the Second Derivative

To determine concavity and inflection points, take the second derivative of the function:First derivative: \[ f'(x) = -4 - 2x \]Second derivative:\[ f''(x) = \frac{d}{dx}(-4 - 2x) = -2 \]

Step 5: Determine Concavity Using Second Derivative

The sign of the second derivative tells us about the concavity:\( f''(x) = -2 \) is constant and negative for all \(x\). This means that the function is always concave down.Since the second derivative does not change sign, there are no intervals where the function is concave up, nor are there any inflection points.

Key Concepts

Understanding Derivatives

Derivatives are an essential concept in calculus. They provide important information about the behavior of functions. The derivative of a function is a measure of how the function's output changes as its input changes. It can be thought of as the function's rate of change at a particular point.

For example, consider the function \(f(x) = 5 - 4x - x^2\). To find the derivative, we apply the rules of differentiation to obtain \(f'(x) = -4 - 2x\). This tells us how the slope of the function changes with respect to \(x\).

• A positive derivative means the function is increasing.
• A negative derivative means the function is decreasing.

By finding derivatives, we can gain insight into the function's increasing and decreasing behavior, which ties closely to the next sections on critical points and intervals.

Concavity Explained

Concavity describes the direction in which a curve opens: upwards or downwards. To determine this, we use the second derivative of the function.

In our example, the second derivative is \(f''(x) = -2\). This indicates a few things:

• If the second derivative is positive, the graph is concave up (U-shape).
• If the second derivative is negative, as in our case, the graph is concave down (n-shape).

In simpler terms, when a graph is concave down, it means as we move along the graph, its slope is getting less steep, or rather, it is curving downward. For \(f(x) = 5 - 4x - x^2\), the concavity is constantly down since \(f''(x) = -2\) for every \(x\).
There are no points where the concavity changes, which means there are no intervals where the function is concave up. Understanding concavity helps in determining points of inflection, which will be discussed next.

Delving into Critical Points

Critical points are values of \(x\) where the first derivative of a function is zero or undefined. These points are essential because they can indicate maxima, minima, or points of inflection.

For \(f(x) = 5 - 4x - x^2\), we found that at \(x = -2\) the first derivative \(f'(x) = -4 - 2x\) equals zero. This marks \(x = -2\) as a critical point. To assess whether this point is a maximum or minimum, we can examine changes in the sign of \(f'(x)\) around it:

• If moving from a positive to a negative sign, it signifies a maximum.
• If moving from a negative to a positive sign, it signifies a minimum.

In our function, test points showed that \(f'(x)\) moved from positive to negative as \(x\) passed through -2, indicating a maximum at that point. Recognizing critical points helps us determine where the function changes its increasing or decreasing nature.

Increasing and Decreasing Intervals

Finding increasing and decreasing intervals requires analyzing the sign of the first derivative over different parts of the function's domain.

In the function \(f(x) = 5 - 4x - x^2\), the critical point \(x = -2\) splits the \(x\)-axis into two intervals: \(( -\infty, -2 )\) and \(( -2, +\infty )\). By picking test points such as \(x = -3\) for the first interval and \(x = 0\) for the second interval, we evaluate \(f'(x)\) to determine the behavior of the function.

• For \(( -\infty, -2 )\), \(f'(-3) = 2\), which indicates that \(f(x)\) is increasing.
• For \(( -2, +\infty )\), \(f'(0) = -4\), showing that \(f(x)\) is decreasing.

Recognizing these intervals where a graph ascends or descends allows for a deeper understanding of the overall shape and behavior of the function. It helps in sketching graphs and predicting function interactions.