Problem 14 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=1-9 x-6 x^{2}-x^{3}$$

Step-by-Step Solution

Answer

Local max at \((-1, 5)\), local min at \((-3, 1)\), inflection point at \((-2, 3)\).

Step 1: Find the derivative

To find the extreme and inflection points, we first need the first and second derivatives of the function. The given function is \( y = 1 - 9x - 6x^2 - x^3 \). The first derivative \( y' \) is found by differentiating:\[ y' = -9 - 12x - 3x^2 \]

Step 2: Find critical points

We set the first derivative equal to zero to find the critical points:\[ -9 - 12x - 3x^2 = 0 \]This can be rewritten as:\[ 3x^2 + 12x + 9 = 0 \]Divide the whole equation by 3 to simplify:\[ x^2 + 4x + 3 = 0 \]Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 4, c = 3 \).\[ x = \frac{-4 \pm \sqrt{16 - 12}}{2} = \frac{-4 \pm 2}{2} \]The solutions are \( x = -1 \) and \( x = -3 \).

Step 3: Determine nature of critical points

To determine the nature of the critical points (local maxima, minima, or neither), we calculate the second derivative. The second derivative \( y'' \) is:\[ y'' = -12 - 6x \]Evaluate \( y'' \) at \( x = -1 \):\[ y''(-1) = -12 - 6(-1) = -12 + 6 = -6 \]Since \( y''(-1) < 0 \), \( x = -1 \) is a local maximum.Evaluate \( y'' \) at \( x = -3 \):\[ y''(-3) = -12 - 6(-3) = -12 + 18 = 6 \]Since \( y''(-3) > 0 \), \( x = -3 \) is a local minimum.

Step 4: Find inflection points

Inflection points occur where \( y'' = 0 \). Set the second derivative to zero:\[ -12 - 6x = 0 \]Solve for \( x \):\[ x = -2 \]At \( x = -2 \), the second derivative changes sign, so there is an inflection point.

Step 5: Determine y-coordinates of found points

Substitute the critical and inflection points back into the original function to find the corresponding \( y \)-coordinates.For \( x = -1 \):\[ y = 1 - 9(-1) - 6(-1)^2 - (-1)^3 = 1 + 9 - 6 + 1 = 5 \]For \( x = -3 \):\[ y = 1 - 9(-3) - 6(-3)^2 - (-3)^3 = 1 + 27 - 54 + 27 = 1 \]For \( x = -2 \):\[ y = 1 - 9(-2) - 6(-2)^2 - (-2)^3 = 1 + 18 - 24 + 8 = 3 \]

Step 6: Graph the function

Plot the points \((-1, 5)\) for the local maximum, \((-3, 1)\) for the local minimum, and \((-2, 3)\) for the inflection point. Then, connect the points considering the behavior of the function in each interval and the nature of the points. The graph should reflect these characteristics and show decreasing, increasing, or concave up/down behavior accordingly.

Key Concepts

Critical Points

Critical points are located where the first derivative of a function equals zero or is undefined. These points are important because they often correspond to local maxima or minima of the function. To find them, we solve the equation \( y' = 0 \). In our function \( y = 1 - 9x - 6x^2 - x^3 \), the derivative is \( y' = -9 - 12x - 3x^2 \). Setting this equal to zero, we simplify it to \( x^2 + 4x + 3 = 0 \). Solving for \( x \), we find two critical points: \( x = -1 \) and \( x = -3 \). These are coordinates where the function potentially changes direction.

Second Derivative Test

The second derivative test helps us determine the nature of critical points found using the first derivative. We calculate the second derivative of our function: \( y'' = -12 - 6x \). By substituting the critical points into this second derivative, we can tell if the point is a local maximum or minimum:

• If \( y''(x) > 0 \), the function is concave up at this point, indicating a local minimum.
• If \( y''(x) < 0 \), the function is concave down, signaling a local maximum.

For \( x = -1 \), \( y'' = -6 \), and since \(-6 < 0\), \( x = -1 \) is a local maximum. For \( x = -3 \), \( y'' = 6 \), hence \( x = -3 \) is a local minimum. This test gives us deeper insight into the graph's critical areas.

Inflection Points

Inflection points occur where the second derivative equals zero and changes sign, indicating a change in concavity. These points don't correspond to maxima or minima but highlight where the graph changes from concave up to concave down, or vice versa. For the function \( y = 1 - 9x - 6x^2 - x^3 \), we solve \( y'' = -12 - 6x = 0 \). This results in \( x = -2 \). Whether \( y'' \) changes sign at this point will determine if it's an actual inflection point. When \( x \) is slightly less than \(-2\), \( y'' \) is positive, and when slightly more, it becomes negative, confirming \( x = -2 \) is an inflection point, making the graph shift its concavity.

Graphing Functions

Graphing functions with the knowledge of critical and inflection points allows for a more accurate representation of the function's behavior. Begin by plotting the critical points \((-1, 5)\) and \((-3, 1)\), along with the inflection point \((-2, 3)\). These points guide the sketching by highlighting regions of increase, decrease, and change in concavity. After plotting, connect these points keeping in mind:

• The intervals where the function is increasing or decreasing, as informed by the first derivative.
• Concavity changes informed by the second derivative and inflection points.

By accurately plotting these points and curves, the graph not only shows maxima and minima but also provides insights into the function's overall behavior, helping visualize the essence of the mathematical concepts taught.