Problem 16 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=1-(x+1)^{3} \end{equation}

Step-by-Step Solution

Answer

The inflection point is at \( x = -1 \). No local or absolute extremum exists.

Step 1: Find the first derivative

To locate critical points, we need to find where the derivative equals zero or is undefined. Start by finding the first derivative of the function \( y = 1 - (x+1)^3 \). The derivative is given by applying the power rule: \( y' = \frac{d}{dx} [1 - (x+1)^3] = -3(x+1)^2 \).

Step 2: Find critical points

Set the first derivative equal to zero to find critical points: \( -3(x+1)^2 = 0 \). Solving for \( x \), we get \( x = -1 \). This is the critical point where the first derivative equals zero.

Step 3: Determine if the critical point is a local extremum

Use the second derivative test to determine the nature of the critical point. First, find the second derivative: \( y'' = \frac{d}{dx} [-3(x+1)^2] = -6(x+1) \). Evaluate the second derivative at the critical point \( x = -1 \): \( y''(-1) = -6(-1+1) = 0 \). Since the second derivative at \( x = -1 \) is zero, the test is inconclusive. Review the behavior around \( x = -1 \) by considering values close to \( x = -1 \) to further analyze.

Step 4: Analyze behavior across other intervals

Pick values around \( x = -1 \) to observe changes in the first derivative: choose \( x = -2 \) and \( x = 0 \). Evaluate \( y'(-2) = -3(-2+1)^2 = -3 \times 1 = -3 \) and \( y'(0) = -3(0+1)^2 = -3 \), confirming \( y'(x) < 0 \) for all \( x \) around the critical point. Hence, there is no local extremum at \( x = -1 \).

Step 5: Check for inflection points

The inflection points occur where the second derivative changes sign. The second derivative is \( y'' = -6(x+1) \), which equals zero at \( x = -1 \). Since \( y'' \) is a linear function of \( x \), it changes from positive for \( x < -1 \) to negative for \( x > -1 \). Therefore, \( x = -1 \) is an inflection point.

Step 6: Graph the function

Graph \( y = 1 - (x+1)^3 \) to visualize its behavior. The graph is a cubic curve with an inflection point at \( x = -1 \). The curve is downward-opening due to the negative leading coefficient of the cubic term. As for extremum points, despite the presence of a critical point, no local or absolute maximum or minimum occurs at \( x = -1 \), confirmed by consistent behavior of \( y'(x) < 0 \) around that point.

Key Concepts

Understanding Critical Points

Critical points are places on the graph of a function where potential peaks or valleys may be located. These points occur where the first derivative of the function is zero or undefined. In simpler terms, you find critical points by calculating the derivative and solving for when it equals zero.

• Critical points can signify potential local maxima or minima.
• If the derivative is undefined at a point, it might also be a critical point.
• In this example, we found the critical point at \(x = -1\).

To determine the nature of the critical point, further testing is necessary with the second derivative or other methods.

Role of the First Derivative

The first derivative represents the rate of change of a function. It is vital in identifying critical points as discussed. To find the first derivative of a function like \(y = 1 - (x+1)^3\), you differentiate using rules of calculus. In this case, the power rule simplifies the process:\[ y' = -3(x+1)^2 \]

• This derivative equals zero at \(x = -1\).
• Negative values indicate the function is decreasing in those intervals.

Thus, around \(x = -1\), the derivative \(y'(x)\) shows a consistent negative value, suggesting no local extremum exists here.

Insights from the Second Derivative

The second derivative offers insight into the concavity of a function and helps confirm the nature of critical points using the second derivative test. For our function, the second derivative is:\[ y'' = -6(x+1) \]This derivative is zero at the same point, \(x = -1\), but it changes signs across intervals:

• Positive before \(x = -1\)
• Negative after \(x = -1\)

Though the test at \(x = -1\) is inconclusive (because \(y''(-1) = 0\)), observing the change of signs suggests important function behavior changes occur at this point, like an inflection.

Exploring Inflection Points

Inflection points are where the curvature of the graph changes, moving from concave up to concave down or vice versa. They occur where the second derivative equals zero and changes sign. In this exercise:

• The second derivative \( y'' = -6(x+1) \) is zero at \( x = -1 \).
• It shifts from positive to negative, confirming \( x = -1 \) as an inflection point.

The curve undergoes a transition at this inflection point, where its bending direction changes, providing crucial insights into the function's graphical behavior.