Problem 76 - Calculus: Early Transcendentals

Question

Sketch the curves. Identify clearly any interesting features, including local maximum and minimum points, inflection points, asymptotes, and intercepts. $$ y=x^{3}+6 x^{2}+9 x $$

Step-by-Step Solution

Answer

Intercepts at (0,0) and (-3,0); local max at (-3,0), local min at (-1,8); inflection at (-2).

Step 1: Identify the function components

The given function is a cubic equation: \[ y = x^3 + 6x^2 + 9x. \] It is a polynomial of degree 3, which implies it is a smooth, continuous curve with no asymptotes but possibly has turning and inflection points.

Step 2: Find the intercepts

To find the y-intercept, set \( x = 0 \):\[ y = 0^3 + 6(0)^2 + 9(0) = 0. \]So the y-intercept is at (0, 0).To find the x-intercepts, set \( y = 0 \):\[ x^3 + 6x^2 + 9x = 0. \]Factor out \( x \):\[ x(x^2 + 6x + 9) = 0. \]This gives \[ x = 0 \] or factoring the quadratic equation further, \( x^2 + 6x + 9 = (x+3)^2 \), so \( x+3 = 0 \). Thus, x-intercepts: \( x = 0 \) and \( x = -3 \).

Step 3: Find the first derivative for critical points

Find the first derivative \( y' \) to identify critical points:\[ y' = 3x^2 + 12x + 9. \]Set \( y' = 0 \) to find critical points:\[ 3x^2 + 12x + 9 = 0. \]Divide by 3:\[ x^2 + 4x + 3 = 0. \]Factor:\[ (x+3)(x+1) = 0. \]Critical points are \( x = -3 \) and \( x = -1 \).

Step 4: Determine local maxima and minima

Evaluate the second derivative \( y''\):\[ y'' = 6x + 12. \]Evaluate \( y'' \) at critical points:- At \( x = -3 \): \[ y''(-3) = 6(-3) + 12 = -18 + 12 = -6. \] Since \( y''(-3) < 0 \), it is a local maximum.- At \( x = -1 \): \[ y''(-1) = 6(-1) + 12 = -6 + 12 = 6. \] Since \( y''(-1) > 0 \), it is a local minimum.

Step 5: Find inflection points

Set \( y'' = 0 \) to determine inflection points:\[ 6x + 12 = 0. \]\[ 6x = -12. \]\[ x = -2. \]Check if the sign of \( y'' \) changes around \( x = -2 \) to confirm it as an inflection point.

Step 6: Sketch the curve

- The intercepts are (0,0) and (-3,0). - Local maximum point at (-3, 0) and local minimum point at (-1, 8). - Inflection point at (-2, some positive y-value). - Sketch the curve ensuring to pass through the intercepts, reach the local max and min, and smoothly transition through the inflection point.

Key Concepts

Cubic Polynomial

A cubic polynomial is a polynomial of degree three, which means its highest power in the equation is three. Such polynomials take the general form:

• \( y = ax^3 + bx^2 + cx + d \)

This type of polynomial typically results in a curve that can have up to two turning points where the graph changes direction.
Cubic polynomials are smooth and continuous with no breaks or asymptotes. Having degree three, they might exhibit interesting features such as critical points and inflection points that make curve sketching a rich exercise.

Critical Points

Critical points occur where the derivative of the function is zero or undefined, marking potential locations for maxima or minima.
In our given function, the first derivative is:

• \( y' = 3x^2 + 12x + 9 \)

Setting this derivative equal to zero helps us find where the function's slope is zero, which is necessary for identifying critical points:

• \( 3x^2 + 12x + 9 = 0 \)
• \( (x+3)(x+1)=0 \)
• Thus, \( x = -3 \) and \( x = -1 \)

These are the x-values where the curve potentially stops increasing and starts decreasing, or vice versa.

Inflection Points

Inflection points are places on the curve where it changes its concavity from concave up to concave down, or vice versa. To find these points, we need to solve:

• \( y'' = 6x + 12\)

Setting the second derivative to zero provides us potential inflection points:

• \( 6x + 12 = 0 \)
• Solving \( 6x = -12 \) gives us \( x = -2 \)

At \( x = -2 \), if the second derivative changes its sign, it confirms an inflection point, meaning the shape of the curve flips from being bowl-shaped upwards to downwards or vice versa.

Local Maximum

A local maximum is a point where the function reaches a high value relative to nearby points. For the function given, we determine a local maximum by evaluating the second derivative at the critical points.

• \( y''(-3) = 6(-3) + 12 = -6 \)

Since this value is negative, it means the curve is concaving downwards, confirming that \(-3\) is a local maximum point.
Thus, there is a peak at \( x = -3 \) where the function ceases to increase and starts decreasing again.

Local Minimum

A local minimum is the lowest point in a localized area of a graph, where the function values are lower than at neighboring points. Calculating at the critical point for \( x = -1 \):

• \( y''(-1) = 6(-1) + 12 = 6 \)

Since this result is positive, the curve has this part concaving upwards, indicating that \(-1\) is a local minimum point.
Here, the function dips to its lowest level before rising again, marking a trough at \( x = -1 \), fittingly known as a local minimum.