Problem 78 - 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

(a) Local max at \(x = 2.1\), local min at \(x = 3.0\). (b) Same x-values; y-values increased by 5.

Step 1: Understanding the Polynomial Function P(x)

The function \(P(x) = (x-1)(x-3)(x-4)\) is a cubic polynomial. Its general form can be expanded to identify the polynomial expression and analyze it.Let's expand it step by step:First, multiply the first two factors: \((x-1)(x-3) = x^2 - 4x + 3\).Now multiply this result by the third factor \(x-4\):\[(x^2 - 4x + 3)(x-4) = x^3 - 4x^2 + 3x - 4x^2 + 16x - 12\]Combine like terms to get the expanded form:\[x^3 - 8x^2 + 19x - 12\]This expression helps us in plotting and analyzing the function.

Step 2: Finding Critical Points of P(x)

To find the local extrema, first find the critical points by taking the derivative of \(P(x)\) and setting it equal to zero:The derivative of \(P(x) = x^3 - 8x^2 + 19x - 12\) is:\[P'(x) = 3x^2 - 16x + 19\].Set the derivative equal to zero to find the critical points:\[3x^2 - 16x + 19 = 0\]Use the quadratic formula to find the roots \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -16\), \(c = 19\).Solving this gives the critical points at approximately \(x_1 = 2.1\) and \(x_2 = 3.0\) (rounded to nearest tenth).

Step 3: Determine Local Extrema of P(x)

To determine whether the critical points are local maxima or minima, evaluate the second derivative, \(P''(x)\):\[P''(x) = 6x - 16\].Evaluate at the critical points:For \(x_1 = 2.1\):\[P''(2.1) = 6(2.1) - 16 = 12.6 - 16 = -3.4\]Since \(P''(2.1) < 0\), \(x_1 = 2.1\) is a local maximum.For \(x_2 = 3.0\):\[P''(3.0) = 6(3.0) - 16 = 18 - 16 = 2\]Since \(P''(3.0) > 0\), \(x_2 = 3.0\) is a local minimum.

Step 4: Graph and Analyze the Function Q(x)

Now consider \(Q(x) = (x-1)(x-3)(x-4) + 5\).This is the same function as \(P(x)\), vertically shifted by 5 units.The local extrema of \(Q(x)\) will be at the same x-values but with their function values increased by 5 units. Therefore:For \(x = 2.1\) (local maximum of \(P(x)\)), the local maximum of \(Q(x)\) is \(P(2.1) + 5\).For \(x = 3.0\) (local minimum of \(P(x)\)), the local minimum of \(Q(x)\) is \(P(3.0) + 5\).

Key Concepts

Local Extrema

Local extrema are the highest or lowest points on a graph within a certain interval. To identify these points on a cubic polynomial like \(P(x)\), we look for points where the graph changes direction. These are typically found at the peaks (local maxima) and valleys (local minima) of the curve.

To determine local extrema mathematically, we need to find the critical points first. Once we have the critical points, the second derivative test helps us decide the nature of each extremum:

• If the second derivative at a critical point is positive, the function is concave upward, indicating a local minimum.
• If the second derivative is negative, the function is concave downward, indicating a local maximum.

This distinction is crucial when interpreting graphs, as it helps to explain the behavior of a function at different intervals.

Critical Points

Critical points of a function occur where the function's derivative equals zero or is undefined. For polynomial functions like \(P(x)\), it is most common to find critical points by setting its derivative equal to zero.

In our example, the derivative of \(P(x)\) is \(P'(x) = 3x^2 - 16x + 19\). Solving \(3x^2 - 16x + 19 = 0\) gives the critical points needed to find local extrema.

Critical points are essential in sketching the shape of the graph because they provide information about where the function changes its increasing or decreasing nature. They essentially serve as the turning points of the polynomial curve.

Polynomial Derivative

Taking the derivative of a polynomial helps us understand the rate and direction of change of the polynomial function. The first derivative of \(P(x)\) provides a new equation, \(P'(x) = 3x^2 - 16x + 19\), which describes the slope of the tangent line to \(P(x)\) at any point \(x\).

By finding where \(P'(x)\) is zero, we're able to locate critical points. These points give us insight into potential local maximums or minimums since they are points where the function's rate of change switches directions.

Derivatives are one of the fundamental tools in calculus that give insight into the geometric characteristics of a graph, showing how a function rises or falls.

Quadratic Formula

The quadratic formula is a reliable method for finding solutions (roots) to quadratic equations of the form \(ax^2 + bx + c = 0\). It is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In the process of finding critical points for \(P(x)\), this formula is instrumental. When we derived \(3x^2 - 16x + 19 = 0\) from the derivative \(P'(x)\), applying the quadratic formula allowed us to pinpoint the critical values \(x_1 = 2.1\) and \(x_2 = 3.0\).

The solutions provided by the quadratic formula are essential in solving polynomial derivative equations, thereby locating critical points for further step-by-step graph analysis.