Problem 12 - Calculus A New Horizon

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. $$f(x)=5+12 x-x^{3}$$

Step-by-Step Solution

Answer

(a) Increasing on \((-2, 2)\), (b) Decreasing on \((-\infty, -2)\) and \((2, \infty)\), (c) Concave up on \((-\infty, 0)\), (d) Concave down on \((0, \infty)\), (e) Inflection point at \(x=0\).

Step 1: Find the first derivative

To determine intervals where the function is increasing or decreasing, we first find the first derivative of the function. Given: \(f(x) = 5 + 12x - x^3\), the first derivative is \(f'(x) = 12 - 3x^2\).

Step 2: Find critical points

Critical points occur where \(f'(x) = 0\) or is undefined. Since \(f'(x) = 12 - 3x^2\), set this equal to zero: \(12 - 3x^2 = 0\). Solving, we get \(x^2 = 4\), so \(x = 2\) and \(x = -2\). These are critical points.

Step 3: Test intervals for increasing/decreasing behavior

Use the critical points \(x = -2\) and \(x = 2\) to divide the number line into intervals: \((-\infty, -2)\), \((-2, 2)\), and \((2, \infty)\). Test each interval by picking a test point and substitute it into \(f'(x)\). For \((-\infty, -2)\), using \(x = -3\), \(f'(-3) = 12 - 3(-3)^2 = -15\), which is negative, so the function decreases here. For \((-2, 2)\), using \(x = 0\), \(f'(0) = 12\), which is positive, so the function increases. For \((2, \infty)\), using \(x = 3\), \(f'(3) = -15\), meaning the function decreases again.

Step 4: Find the second derivative

To determine intervals of concavity, find the second derivative of the function. \(f''(x) = -6x\).

Step 5: Find possible inflection points

Inflection points occur where \(f''(x) = 0\). Set \(-6x = 0\), giving \(x = 0\). Check changes in concavity at this point.

Step 6: Test intervals for concavity

Use \(x = 0\) to split the number line into intervals \((-\infty, 0)\) and \((0, \infty)\). Test these intervals using \(f''(x)\). For \((-\infty, 0)\), using \(x = -1\), \(f''(-1) = 6\), so it's concave up. For \((0, \infty)\), using \(x = 1\), \(f''(1) = -6\), so it's concave down.

Key Concepts

Critical Points

Critical points are crucial in calculus as they help us understand where a function changes its increasing or decreasing behavior. A critical point occurs when the first derivative of the function is equal to zero or is undefined.

Let's break it down with our exercise. For the function \( f(x) = 5 + 12x - x^3 \), we first find its first derivative: \( f'(x) = 12 - 3x^2 \). Now we set \( f'(x) = 0 \) to identify critical points. This gives us the equation \( 12 - 3x^2 = 0 \). Solving this for \( x \), we get \( x = 2 \) and \( x = -2 \).

These \( x \)-values are the critical points of the function, indicating where the behavior of the function might change.

Derivative

The derivative is a fundamental tool in calculus, often used to determine the rate at which a function is changing. In this context, the first derivative helps us find where our function is increasing or decreasing.

For our function \( f(x) = 5 + 12x - x^3 \), its first derivative is \( f'(x) = 12 - 3x^2 \). By analyzing this derivative, we can find:

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

We tested intervals around the critical points \( x = -2 \) and \( x = 2 \) and found that the function decreases on \((-finity, -2)\) and \((2, finity)\), and increases on \((-2, 2)\).

This analysis helps in sketching the graph and understanding the behavior of \( f(x) \).

Concavity

Concavity tells us whether a function curves upwards or downwards. To determine concavity, we use the second derivative.

For \( f(x) = 5 + 12x - x^3 \), we find the second derivative: \( f''(x) = -6x \). We analyze this by choosing test points in intervals defined around points where \( f''(x) = 0 \). Here's how to interpret it:

• If \( f''(x) > 0 \), the graph is concave up (shaped like \( \cup \)).
• If \( f''(x) < 0 \), the graph is concave down (shaped like \( \cap \)).

We found \( x = 0 \) as a potential inflection point, splitting the number line into \((-finity, 0) \text{ and } (0, finity)\). Testing these intervals showed concavity up on \((-finity, 0)\) and concavity down on \((0, finity)\). This information helps us understand how the slope of \( f(x) \) changes.

Inflection Points

Inflection points are where the curvature of a graph changes from concave up to concave down, or vice versa.

To find an inflection point, set the second derivative \( f''(x) \) to zero. For our function, \( f''(x) = -6x \), leading us to find \( x = 0 \) as a potential point of inflection.

At \( x = 0 \), there's a change in the concavity:

• The function is concave up before \( x = 0 \) (for \( x < 0 \)).
• The function is concave down after \( x = 0 \) (for \( x > 0 \)).

This confirms that \( x = 0 \) is indeed an inflection point. Inflection points give us critical insights into how the shape and direction of the graph change.