Problem 74 - Algebra and Trigonometry

Question

(a) Graph the function \(P(x)=(x-1)(x-3)(x-4)\) and find all local extrema, correct to the nearest tenth. (b) Graph the function $$Q(x)=(x-1)(x-3)(x-4)+5$$ and use your answers to part (a) to find all local extrema, correct to the nearest tenth.

Step-by-Step Solution

Answer

For P(x), local extrema at x ≈ 1.8 and x ≈ 3.1. For Q(x), local extrema occur at same x-values, y-values are P(x) + 5.

Step 1: Define the Function P(x)

The function given is \( P(x) = (x-1)(x-3)(x-4) \). This is a polynomial of degree 3, which is a cubic function.

Step 2: Identify Critical Points for P(x)

To find critical points, we need to take the derivative of \( P(x) \), set it equal to zero, and solve for \( x \). Calculate \( P'(x) \), which involves using the product rule for the three factors.

Step 3: Calculate Derivative of P(x)

First, expand the product to simplify differentiation: \( P(x) = (x^2 - 4x + 3)(x - 4) = x^3 - 9x^2 + 16x - 12 \). Now, differentiate to find \( P'(x) = 3x^2 - 18x + 16 \).

Step 4: Solve for Critical Points of P(x)

Set \( P'(x) = 0 \) to find critical points: \( 3x^2 - 18x + 16 = 0 \). Solve this quadratic equation using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 3, b = -18, c = 16 \).

Step 5: Calculate Critical Points for P(x)

Substitute into the quadratic formula to get \( x = \frac{18 \pm \sqrt{324 - 192}}{6} \), which simplifies to \( x = \frac{18 \pm \sqrt{132}}{6} \). Simplify further to get \( x \approx 1.8 \) and \( x \approx 3.1 \).

Step 6: Find Local Extrema for P(x)

Analyze the sign changes of \( P'(x) \) around the critical points \( x = 1.8 \) and \( x = 3.1 \) to determine extrema. Check that \( P'(x) \) changes sign at these points confirming they are local extrema.

Step 7: Define the Function Q(x)

For \( Q(x) = (x-1)(x-3)(x-4) + 5 \), notice it's just \( P(x) + 5 \). This transformation is a vertical shift of 5 units upward of \( P(x) \).

Step 8: Find Local Extrema for Q(x)

The local extrema for \( Q(x) \) are the same x-values as \( P(x) \), since the vertical shift does not change the x-locations of extrema, but the new y-values are \( P(x) + 5 \). Calculate \( Q(1.8) \) and \( Q(3.1) \) based on values from \( P(x) \).

Step 9: Local Extrema for Q(x)

Obtain \( Q(1.8) \approx P(1.8) + 5 \) and \( Q(3.1) \approx P(3.1) + 5 \) using the critical points found; these are the local extrema of \( Q(x) \).

Key Concepts

Cubic Functions

Cubic functions are a type of polynomial function where the highest degree of the variable is three. A general cubic function has the form \( ax^3 + bx^2 + cx + d \). These functions are characterized by their potential to have up to three roots, two turning points, and variable end behavior dependent on the leading coefficient.

For the function \( P(x) = (x-1)(x-3)(x-4) \), it is structured in factored form, showcasing its roots or x-intercepts: \( x = 1, 3, \) and \( 4 \). Once expanded, it takes on the traditional cubic form.

• Cubic functions can exhibit unique shapes like the classic 'S' curve, due to these turning points.
• They often cross the x-axis one or three times based on their real roots.
• Understanding the structure of cubic functions helps predict their general behavior and appearance on a graph.

Critical Points

Critical points are where a function's derivative is zero or undefined, indicating potential maximums, minimums, or inflection points. For the function \( P(x) = x^3 - 9x^2 + 16x - 12 \), we calculate the derivative \( P'(x) = 3x^2 - 18x + 16 \) and set it to zero to find these critical points.

To solve \( 3x^2 - 18x + 16 = 0 \), use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), resulting in \( x \approx 1.8 \) and \( x \approx 3.1 \).

• These points are vital because they usually signify changes in the graph's direction.
• By finding where a derivative equals zero, we determine where the slope of the tangent to the curve is horizontal.
• These points help us pinpoint where the curve might "turn around" or change its trend.

Local Extrema

Local extrema of a function refer to its local maximum or minimum values. They occur at critical points under the right conditions, where the derivative changes sign. For the function \( P(x) = x^3 - 9x^2 + 16x - 12 \), you can find local extrema by evaluating \( P(x) \) at the critical points \( x = 1.8 \) and \( x = 3.1 \).

An extremum is a local minimum if the function moves from decreasing to increasing, and a local maximum if the movement is from increasing to decreasing.

• Local maxima are peaks on a graph whereas local minima are troughs.
• Evaluating \( P(x) \) at these critical values gives insight into the specific height or depth of the graph at these points.
• Understanding these extrema helps in sketching accurate graphs and interpreting the function's behavior over intervals.

Derivative

The derivative of a function provides critical information about its rate of change and behavior, especially near critical points. In calculus, the derivative of a polynomial like \( P(x) = x^3 - 9x^2 + 16x - 12 \) is calculated to find slopes of tangents across different points on the graph.

For the given function, the derivative \( P'(x) = 3x^2 - 18x + 16 \) informs us where the function is increasing, decreasing, or holding a constant rate of change.

• The derivative highlights points of inflection where the graph changes from concave up to concave down, or vice versa.
• By setting derivative equal to zero, we can locate critical points, assisting in plotting and understanding the curve's overall shape.
• Derivatives are essential tools for modeling and predicting real-world phenomena represented by polynomial functions.