Problem 11 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

Local maximum at \((-1, 5)\), local minimum at \((1, 1)\), and inflection point at \((0, 3)\).

Step 1: Find the First Derivative

To determine local extreme points, we first need to find the critical points of the function by calculating its first derivative. The function given is \( y = x^3 - 3x + 3 \). The derivative using power rule is: \[ y' = 3x^2 - 3. \]

Step 2: Set First Derivative to Zero

To find the critical points, we set the first derivative equal to zero and solve for \( x \): \[ 3x^2 - 3 = 0. \]Factor the equation:\[ 3(x^2 - 1) = 0. \]Solve for \( x \): \( x^2 = 1 \), hence \( x = 1 \) or \( x = -1 \).

Step 3: Determine Local Extrema Using Second Derivative

To determine whether these critical points are local maxima, minima, or saddle points, we find the second derivative and use the second derivative test. The second derivative of the function is:\[ y'' = 6x. \]Evaluate the second derivative at the critical points:- At \( x = 1 \), \( y'' = 6(1) = 6 > 0 \), indicates a local minimum.- At \( x = -1 \), \( y'' = 6(-1) = -6 < 0 \), indicates a local maximum.

Step 4: Identify Any Inflection Points

An inflection point occurs where the second derivative changes sign. Solve \( y'' = 6x = 0 \) to find potential inflection points.\[ x = 0. \]Since the second derivative changes sign at \( x = 0 \) (from negative to positive), this point is an inflection point.

Step 5: Determine Absolute Extreme and Graphical Interpretation

Since \( y = x^3 - 3x + 3 \) is a cubic function, it extends to infinity in both positive and negative directions. Therefore, it does not have absolute extrema. Plotting the graph, observe the turning points at local maxima \((-1, 5)\) and minima \((1, 1)\), and the inflection point at \((0, 3)\).

Step 6: Graph Function

The graph of \( y = x^3 - 3x + 3 \) shows a local maximum at \( x = -1 \), local minimum at \( x = 1 \), and an inflection point at \( x = 0 \). The overall shape confirms this cubic function has no absolute extrema.

Key Concepts

First Derivative

The first derivative of a function is crucial in identifying where the slope of the tangent to the curve is zero, which often leads to finding critical points. For the function given, \( y = x^3 - 3x + 3 \), the first derivative is found using the power rule.\[y' = 3x^2 - 3. \]This expression tells us how \( y \) changes with respect to \( x \). By setting this derivative equal to zero:

• \( 3x^2 - 3 = 0 \)

We can solve for \( x \) to find where the slope is zero, indicating potential local extrema.

Second Derivative

The second derivative of a function helps us determine the concavity of the function, which is used in the second derivative test to classify critical points. For the function:\[ y = x^3 - 3x + 3, \]the second derivative is calculated as:\[ y'' = 6x. \]By evaluating this second derivative at critical points, we determine whether these points are maxima or minima:

• At \( x = 1 \), \( y'' = 6 \times 1 = 6 \), which is positive, indicating concave up and thus a local minimum.
• At \( x = -1 \), \( y'' = 6 \times (-1) = -6 \), which is negative, indicating concave down and thus a local maximum.

Inflection Point

An inflection point is where the concavity of the function changes, indicated by a sign change in the second derivative. For the function's second derivative, \( y'' = 6x \), an inflection point occurs when:

• \( 6x = 0 \)

Solving for \( x \), we find \( x = 0 \). At this point, the second derivative changes from negative (concave down) to positive (concave up), confirming the inflection point at \( x = 0 \). This shift in concavity is key in understanding the behavior of the function.

Local Extrema

Local extrema are found at critical points where a function changes direction, indicating local maxima or minima. In this exercise:

• \( x = -1 \) is a local maximum because \( y'' = -6 \) (concave down).
• \( x = 1 \) is a local minimum because \( y'' = 6 \) (concave up).

These points represent peaks and valleys in the graph of the function, but do not represent overall highest or lowest points, since cubic functions extend indefinitely.

Critical Points in Calculus

Critical points are where the first derivative of a function is zero or undefined, marking potential locations for local extrema. By setting the first derivative equal to zero:

• \( 3x^2 - 3 = 0 \)
• \( x = 1 \) and \( x = -1 \)

We'll find where the function may have turning points or change direction relative to the original. These critical points are invaluable in analyzing the overall shape and features of the function's graph.