Problem 9 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=2 x^{3}-3 x^{2}-12 x+1 $$

Step-by-Step Solution

Answer

The function is concave up on \((\frac{1}{2}, \infty)\), concave down on \((-\infty, \frac{1}{2})\), with an inflection point at \( x = \frac{1}{2} \). Local maximum at \( x = -1 \) and local minimum at \( x = 2 \).

Step 1: Find the First Derivative

To start, find the first derivative of the function \( f(x) = 2x^3 - 3x^2 - 12x + 1 \). This will help us find the critical points. The first derivative is found using power rules: \( f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 1) = 6x^2 - 6x - 12 \).

Step 2: Find Critical Points

Find the critical points by setting the first derivative equal to zero: \( 6x^2 - 6x - 12 = 0 \). Simplify and solve for \( x \): \( x^2 - x - 2 = 0 \). Factoring gives \((x-2)(x+1) = 0\), so \( x = 2 \) and \( x = -1 \). These are the critical points.

Step 3: Find the Second Derivative

Compute the second derivative of \( f(x) \) as follows: \( f''(x) = \frac{d}{dx}(6x^2 - 6x - 12) = 12x - 6 \). The second derivative will help determine concavity and evaluate critical points.

Step 4: Determine Concavity

Find intervals of concavity using the second derivative. Set \( f''(x) = 0 \) to find potential inflection points: \( 12x - 6 = 0 \) gives \( x = \frac{1}{2} \). Test intervals around \( x = \frac{1}{2} \) to determine concavity. For \( x < \frac{1}{2} \), select \( x = 0 \): \( f''(0) = -6 \), concave down; for \( x > \frac{1}{2} \), select \( x = 1 \): \( f''(1) = 6 \), concave up.

Step 5: Find Points of Inflection

The points of inflection occur where the concavity changes from down to up or up to down. Since the concavity changes at \( x = \frac{1}{2} \), there is a point of inflection at \( x = \frac{1}{2} \).

Step 6: Apply the Second Derivative Test

Apply the second derivative test to the critical points found in Step 2. For \( x = -1 \), \( f''(-1) = 12(-1) - 6 = -18 \), indicating a local maximum. For \( x = 2 \), \( f''(2) = 12(2) - 6 = 18 \), indicating a local minimum.

Key Concepts

Critical Points

Critical points of a function occur where the first derivative is zero or undefined. This is because at critical points, the slope of the tangent (given by the first derivative) is level, meaning it is neither increasing nor decreasing. Thus, finding critical points is an essential first step in analyzing any function's behavior.
To find critical points for the function provided, we start by taking the derivative to find the slope. The derivative of the function is given by

• \( f'(x) = 6x^2 - 6x - 12 \)

Next, set the first derivative equal to zero to find where the slope levels off:

• \( 6x^2 - 6x - 12 = 0 \)

This equation can be simplified and factored as follows:

• \( x^2 - x - 2 = 0 \)
• \( (x-2)(x+1) = 0 \)
• \( x = 2 \) and \( x = -1 \)

Therefore, the critical points occur at \( x = 2 \) and \( x = -1 \). At these points, the function changes its rate of change, paving the way for further analysis using the second derivative test.

Concavity

Concavity reveals whether a curve is shaped upwards like a cup or downwards like a cap. To determine concavity for the function, we utilize the second derivative.

• The second derivative is found by differentiating the first derivative: \( f''(x) = 12x - 6 \)

Check where the second derivative is zero to identify potential points where concavity changes:

• Set \( f''(x) = 0 \): \( 12x - 6 = 0 \)
• Solve for \( x \): \( x = \frac{1}{2} \)

To determine where the function is concave up or down, test intervals on either side of \( x = \frac{1}{2} \):

• For \( x < \frac{1}{2} \) (choose \( x = 0 \)): \( f''(0) = -6 \), therefore the function is concave down.
• For \( x > \frac{1}{2} \) (choose \( x = 1 \)): \( f''(1) = 6 \), so the function is concave up.

This analysis informs us that the curve is concave down for \( x < \frac{1}{2} \) and concave up for \( x > \frac{1}{2} \). Understanding these intervals of concavity is fundamental once moving towards identifying points of inflection.

Points of Inflection

A point of inflection is where a graph changes its concavity – going from concave up to concave down or vice versa. Such a shift indicates a fundamental change in the curve's shape.
The examination of the concavity showed that the only change occurs at \( x = \frac{1}{2} \). This is confirmed by:

• At \( x = \frac{1}{2} \), the concavity of the function shifts from down to up.

Thus, an inflection point exists at \( x = \frac{1}{2} \). Inflection points are noteworthy because they signal shifts from acceleration to deceleration in the graph’s curvature. Capturing these shifts help in beautifully sketching the behavior of the overall function.

Local Maximum and Minimum

Local maxima and minima refer to the highest or lowest points in a particular section of the graph. Finding these points helps us understand peaks and troughs within intervals of the function.
To determine whether the critical points at \( x = -1 \) and \( x = 2 \) represent a local maximum or minimum, the second derivative test is deployed:

• For \( x = -1 \), calculate second derivative: \( f''(-1) = 12(-1) - 6 = -18 \). Since this is negative, \( x = -1 \) is a local maximum.
• For \( x = 2 \), calculate second derivative: \( f''(2) = 12(2) - 6 = 18 \). As this is positive, \( x = 2 \) is a local minimum.

These results illustrate where the function reaches high points and dives into low points locally. Identifying these helps especially in optimization problems or when modeling naturally occurring phenomena.