Problem 57 - Precalculus Mathematics for Calculus

Question

Find the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. $$g(x)=x^{4}-2 x^{3}-11 x^{2}$$

Step-by-Step Solution

Answer

Local maxima at \(x = 0\) and \(x = -1.71\) with values 0 and 30.07; local minimum at \(x = 3.21\) with value -51.27.

Step 1: Determine the first derivative

To find the critical points where local maxima or minima may occur, we first need to find the derivative of the function. The function is given by:\[ g(x) = x^4 - 2x^3 - 11x^2 \]Differentiating with respect to \(x\), we get:\[ g'(x) = 4x^3 - 6x^2 - 22x \]

Step 2: Find critical points

Critical points occur where the first derivative is zero or undefined. Set \(g'(x) = 0\):\[ 4x^3 - 6x^2 - 22x = 0 \]Factor the equation:\[ x(4x^2 - 6x - 22) = 0 \]This gives us one critical point at \(x = 0\). We further solve the quadratic equation:\[ 4x^2 - 6x - 22 = 0 \]

Step 3: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 4\), \(b = -6\), \(c = -22\), we solve for \(x\):\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 4 \times (-22)}}{2 \times 4} \]\[ x = \frac{6 \pm \sqrt{36 + 352}}{8} \]\[ x = \frac{6 \pm \sqrt{388}}{8} \]\[ x = \frac{6 \pm 19.70}{8} \]Calculating the solutions:\[ x_1 = \frac{25.70}{8} = 3.21 \quad \text{(rounded to two decimal places)} \]\[ x_2 = \frac{-13.70}{8} = -1.71 \quad \text{(rounded to two decimal places)} \]

Step 4: Determine local extrema using the second derivative test

We use the second derivative to determine the nature of the critical points. Compute the second derivative:\[ g''(x) = 12x^2 - 12x - 22 \]Evaluate \(g''(x)\) at the critical points derived:At \(x = 0\):\[ g''(0) = -22 \]Since \(g''(0) < 0\), \(x = 0\) is a local maximum.At \(x = 3.21\):\[ g''(3.21) = 12(3.21)^2 - 12(3.21) - 22 = 72 \]Since \(g''(3.21) > 0\), \(x = 3.21\) is a local minimum.At \(x = -1.71\):\[ g''(-1.71) = 12(-1.71)^2 - 12(-1.71) - 22 = -26.38 \]Since \(g''(-1.71) < 0\), \(x = -1.71\) is a local maximum.

Step 5: Calculate function values at critical points

Finally, substitute the x-values back into the original function \(g(x)\) to find the function values at each critical point:For \(x = 0\):\[ g(0) = 0^4 - 2 \times 0^3 - 11 \times 0^2 = 0 \]For \(x = 3.21\):\[ g(3.21) \approx (3.21)^4 - 2(3.21)^3 - 11(3.21)^2 \approx -51.27 \]For \(x = -1.71\):\[ g(-1.71) \approx (-1.71)^4 - 2(-1.71)^3 - 11(-1.71)^2 \approx 30.07 \]

Key Concepts

Critical Points

A critical point in the context of calculus is where the derivative of a function is either zero or undefined. These points are important because they are potential points where the function might have a local maximum or minimum. To find critical points, you first take the derivative of the function. For our function, the derivative is:

• \( g'(x) = 4x^3 - 6x^2 - 22x \)

The next step is to set this derivative equal to zero and solve for \(x\):

• \( 4x^3 - 6x^2 - 22x = 0 \)

Factoring out \(x\), the equation becomes:

• \( x(4x^2 - 6x - 22) = 0 \)

This immediately gives \(x = 0\) as a critical point. For the quadratic part, you use the quadratic formula:

• \( x = \frac{6 \pm \sqrt{36 + 352}}{8} \)


• This yields critical points at \(x = 3.21\) and \(x = -1.71\).

Identifying critical points sets the stage for determining whether these are points of maxima or minima.

First Derivative

The first derivative of a function represents the rate of change or the slope of the function at any given point. By finding where this derivative equals zero or is undefined, we can identify critical points. In our function, the first derivative is:

• \( g'(x) = 4x^3 - 6x^2 - 22x \)

Taking the derivative is the initial step in locating potential locations for local maxima or minima.
Setting the derivative \(g'(x)\) to zero helps to pinpoint where the function’s slope is flat, indicating a possible extremum:

• \( 4x^3 - 6x^2 - 22x = 0 \)

By factoring out \(x\), we note any solution to this equation might indicate a critical point:

• \(0, 3.21, -1.71\)

The first derivative test also helps in confirming the behavior of the function around critical points, though, in this exercise, we proceed directly to the second derivative for this purpose.

Second Derivative Test

The second derivative test is a method used to classify the critical points identified by the first derivative test. It involves taking the second derivative of the function and evaluating it at each critical point.

• For our function, the second derivative is: \( g''(x) = 12x^2 - 12x - 22 \)

By substituting the critical points into the second derivative, we determine whether each point is a local maximum, minimum, or a saddle point. The rules are:

• If \( g''(x) > 0\) at a critical point, the function has a local minimum there.
• If \( g''(x) < 0\) at a critical point, the function has a local maximum there.
• If \( g''(x) = 0\), the test is inconclusive.

For the critical points:

• At \( x = 0\), \( g''(0) = -22 \), which indicates a local maximum.
• At \( x = 3.21\), \( g''(3.21) = 72 \), indicating a local minimum.
• At \( x = -1.71\), \( g''(-1.71) = -26.38 \), which also indicates a local maximum.

Understanding the second derivative allows you to classify the function's critical points, providing a fuller picture of the function's shape.