Problem 18 - 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. $$f(x)=5+12 x-x^{3}$$

Step-by-Step Solution

Answer

(a) \((-2, 2)\); (b) \((-\infty, -2)\), \((2, \infty)\); (c) \((-\infty, 0)\); (d) \((0, \infty)\); (e) \( x = 0 \).

Step 1: Find the First Derivative

Begin by finding the first derivative of the function. Given the function \( f(x) = 5 + 12x - x^3 \), the first derivative is \( f'(x) = 12 - 3x^2 \). This derivative will help us determine where the function is increasing or decreasing.

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find the critical points: \( 12 - 3x^2 = 0 \). Solving this gives \( x^2 = 4 \), so \( x = 2 \) and \( x = -2 \). These are potential points where the function changes from increasing to decreasing or vice versa.

Step 3: Test Intervals for Increasing/Decreasing

Using the critical points \(-2\) and \(2\), divide the number line into intervals: \((-\infty, -2)\), \((-2, 2)\), and \((2, \infty)\). Test a point within each interval to determine if \( f'(x) \) is positive (increasing) or negative (decreasing). The test points can be \( x = -3, 0, \text{and } 3 \). - For \( x = -3 \), \( f'(-3) = 12 - 3(-3)^2 = -15 \) (decreasing).- For \( x = 0 \), \( f'(0) = 12 \) (increasing).- For \( x = 3 \), \( f'(3) = 12 - 3(3)^2 = -15 \) (decreasing). So, \( f \) is increasing on \((-2, 2)\) and decreasing on \((-\infty, -2)\) and \((2, \infty)\).

Step 4: Find the Second Derivative

Find the second derivative to analyze concavity: \( f''(x) = -6x \). This will help us determine which intervals are concave up or down.

Step 5: Solve for Inflection Points

Set the second derivative equal to zero to find potential inflection points: \( -6x = 0 \). This gives \( x = 0 \). Test intervals around this point to determine concavity changes.

Step 6: Test Intervals for Concavity

Using the potential inflection point \( x = 0 \), divide the number line into intervals: \((-\infty, 0)\) and \((0, \infty)\). - For \( x = -1 \), \( f''(-1) = 6 \) (concave up).- For \( x = 1 \), \( f''(1) = -6 \) (concave down). So, \( f \) is concave up on \((-\infty, 0)\) and concave down on \((0, \infty)\).

Step 7: Inflection Point

Identify the inflection point as the point where the concavity changes, which occurs when \( x = 0 \). Hence, the inflection point is at \( x = 0 \).

Key Concepts

Function Increasing Decreasing Intervals

When we explore the intervals where a given function increases or decreases, we're essentially studying how its rate of change behaves across different segments of its domain.
By calculating the first derivative of a function, denoted as \( f'(x) \), we can pinpoint these transitions.Here's a simple rundown of the steps:

• First, find the derivative \( f'(x) \) of the function. For example, for the function \( f(x) = 5 + 12x - x^3 \), the first derivative is \( f'(x) = 12 - 3x^2 \).
• Identify critical points by setting \( f'(x) = 0 \). These points are where the function changes direction from increasing to decreasing or vice versa. Solving \( 12 - 3x^2 = 0 \) yields the solutions \( x = 2 \) and \( x = -2 \).
• Divide the number line into intervals using these critical points and test a value from each to determine if the function is increasing or decreasing based on the sign of \( f'(x) \).

For our example:- \( f(x) \) increases in the interval \((-2, 2)\) where \( f'(x) > 0 \).- \( f(x) \) decreases in the intervals \((-fty, -2)\) and \((2, \infty)\) where \( f'(x) < 0 \).

Concavity

Concavity tells us about the direction a curve bends at different intervals. It is the systemic curve behavior dictated by the second derivative of a function.
With its help, we can determine if the function is bending upwards or downwards.Here's what you need to do:

• First, calculate the second derivative, denoted \( f''(x) \). For our function, the second derivative is \( f''(x) = -6x \).
• If \( f''(x) > 0 \), the function is concave up (like a cup opening upwards), and if \( f''(x) < 0 \), it's concave down.
• Check the sign of \( f''(x) \) within various intervals to determine concavity.

For \( f(x) = 5 + 12x - x^3 \):- It's concave up on \((-\infty, 0)\).- It's concave down on \((0, \infty)\).

Inflection Points

Inflection points are where a function's concavity changes from up to down or from down to up.
These are critical to understanding how the entire function shifts its curve behavior.Here is how to find them:

• Find where \( f''(x) = 0 \). These are potential inflection points.
• Verify changes in concavity around these points by checking the sign of \( f''(x) \).
• Concretely, a change in sign confirms an inflection point.

For our example function:- Solve \( -6x = 0 \), which gives \( x = 0 \) as a potential inflection point.- Testing intervals to the left and right of \( x = 0 \), we find a change in concavity, confirming \( x = 0 \) is indeed an inflection point.