Problem 2 - Calculus A New Horizon

Question

Give a complete graph of the polynomial, and label the coordinates of the stationary points and inflection points. Check your work with a graphing utility. $$x^{3}-3 x+1$$

Step-by-Step Solution

Answer

Stationary points: \((1, -1), (-1, 3)\); Inflection point: \((0, 1)\).

Step 1: Find the derivative of the polynomial

To find the stationary points, we need the first derivative of the polynomial. The function is \( f(x) = x^3 - 3x + 1 \). The derivative \( f'(x) \) is calculated by applying basic differentiation rules: \( f'(x) = 3x^2 - 3 \).

Step 2: Solve for stationary points

Set the derivative equal to zero to find the stationary points: \( 3x^2 - 3 = 0 \). Solving this gives \( 3x^2 = 3 \) which simplifies to \( x^2 = 1 \). Thus, \( x = \pm 1 \).

Step 3: Calculate y-values for stationary points

Substitute the x-values back into the original equation to find their corresponding y-values. For \( x = 1 \), \( y = 1^3 - 3 \times 1 + 1 = -1 \). For \( x = -1 \), \( y = (-1)^3 - 3 \times (-1) + 1 = 3 \). So, the stationary points are \( (1, -1) \) and \( (-1, 3) \).

Step 4: Find the second derivative for inflection points

To find inflection points, calculate the second derivative \( f''(x) \). From \( f'(x) = 3x^2 - 3 \), differentiate again to get \( f''(x) = 6x \).

Step 5: Solve for inflection points

Set the second derivative to zero: \( 6x = 0 \). Solving this gives \( x = 0 \).

Step 6: Calculate the y-value for inflection point

Substitute \( x = 0 \) into the original equation: \( y = 0^3 - 3 \times 0 + 1 = 1 \). Thus, the inflection point is \( (0, 1) \).

Step 7: Graph the polynomial and label critical points

Sketch the graph of \( f(x) = x^3 - 3x + 1 \). The polynomial graph should display a local maximum at \( (-1, 3) \) and a local minimum at \( (1, -1) \), with an inflection point at \( (0, 1) \).

Step 8: Verify with a graphing utility

Use a graphing utility to confirm that the plotted graph accurately reflects the stationary points and inflection point calculated. The points \( (1, -1) \), \( (-1, 3) \), and \( (0, 1) \) should appear at the corresponding positions on the graph.

Key Concepts

Stationary Points

Stationary points on a graph are where the slope of the curve is zero. These occur at points where the derivative of the function equals zero. For the polynomial function \( f(x) = x^3 - 3x + 1 \), we find the first derivative to identify these points. The derivative is calculated as \( f'(x) = 3x^2 - 3 \).

Next, set the derivative equal to zero: \( 3x^2 - 3 = 0 \). Solving gives \( x^2 = 1 \), which means \( x = \pm 1 \). These \( x \)-values are where potential stationary points exist. To confirm these are indeed stationary points, plug them back into the original function to find the corresponding \( y \)-values.

• For \( x = 1 \): \( y = 1^3 - 3 \times 1 + 1 = -1 \). Hence, \( (1, -1) \) is a stationary point.
• For \( x = -1 \): \( y = (-1)^3 - 3 \times (-1) + 1 = 3 \). Thus, \( (-1, 3) \) is another stationary point.

These points indicate where the graph either reaches local minimums or maximums.

Inflection Points

Inflection points are points on a curve where the concavity changes, typically identified by the second derivative of the function. For our polynomial \( f(x) = x^3 - 3x + 1 \), we first find the second derivative. Begin with the first derivative \( f'(x) = 3x^2 - 3 \), then calculate the second derivative: \( f''(x) = 6x \).

To determine the inflection point, set \( f''(x) \) to zero: \( 6x = 0 \). Solving this yields \( x = 0 \), suggesting a potential inflection point at this \( x \)-value. To confirm, substitute \( x = 0 \) back into the original function to calculate the \( y \)-coordinate: \( y = 0^3 - 3\times 0 + 1 = 1 \). Thus, the inflection point is \( (0, 1) \).

At this point, the graph transitions from concave up to concave down or vice versa. Verifying this shift can help ensure the graph's accuracy when plotted.

Derivative Calculations

Calculating derivatives is crucial for determining the stationary and inflection points discussed earlier. A derivative essentially measures how a function's output value changes as its input changes. For the function \( f(x) = x^3 - 3x + 1 \), the first derivative \( f'(x) = 3x^2 - 3 \) provides the slope of the tangent to the curve at any given point. This first derivative helps identify where the curve has horizontal tangents, i.e., stationary points.

The second derivative, \( f''(x) = 6x \), shows how the slope itself changes, indicating the curvature of the graph. When this derivative equals zero, it suggests inflection points where the concavity changes.

• Derivatives are calculated using power rules and chain rules, which simplify complex polynomial functions.
• These calculations are pivotal for accurately sketching the graph of any polynomial function, helping in determining critical points like where the graph peaks, troughs, or inflects.

Understanding derivative calculations lays a robust foundation for studying calculus and analyzing polynomial functions effectively.