Problem 60 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the relative maxima and relative minima, if any, of each function. $$ f(x)=x^{3}-3 x+6 $$

Step-by-Step Solution

Answer

The function \(f(x)=x^3-3x+6\) has a relative maximum at \(x=-1\) and a relative minimum at \(x=1\).

Step 1: Find the first derivative of the function

To find the first derivative of the function, we will apply the power rule to each term of the function: \(f'(x) = \frac{d}{dx}(x^3 - 3x + 6)\) \(f'(x) = 3x^2 - 3\)

Step 2: Find the critical points

To find the critical points, set the first derivative equal to zero and solve for x: \(3x^2 - 3 = 0\) This equation can be factored: \(3(x^2 - 1) = 0\) So, \(x^2 - 1 = 0\) and \(x^2=1\) Solving for x, we get two critical points: \(x= \pm 1\)

Step 3: Find the second derivative of the function

Taking the derivative of the first derivative, we can find the second derivative: \(f''(x)= \frac{d^2}{dx^2}(3x^2-3)\) \(f''(x)=6x\)

Step 4: Use the second derivative test to classify critical points

Plug the critical points into the second derivative to determine if they are relative maxima or relative minima: For \(x=1\), \(f''(1)=6(1)=6 > 0\), which indicates a relative minimum at \(x=1\). For \(x=-1\), \(f''(-1)=6(-1)=-6 < 0\), which indicates a relative maximum at \(x=-1\). So, we have found a relative maximum at \(x=-1\) and a relative minimum at \(x=1\).

Key Concepts

Critical Points

In the journey to determine the hills and valleys of a function, locating the critical points is the first crucial step. These are the spots where the function's slope is zero or where the derivative doesn't exist.
For the function f(x) = x^3 - 3x + 6, by finding the first derivative and setting it to zero, we pinpoint the critical points. Mathematically, this translates to solving f'(x) = 3x^2 - 3 = 0, leading us to the solutions x = ±1. These points are like the coordinates where the function takes a breath, either peaking or troughing, hence they can potentially be the relative extrema we seek.

First Derivative Test

The First Derivative Test is a handy technique to analyze the behavior of a function around its critical points. After honing in on the critical points x = ±1, this test involves examining the sign of the derivative before and after each point. It's like checking the function’s mood; if the derivative switches from positive to negative, we're at the top of a hill (a relative maximum). Conversely, if the sign flips from negative to positive, we're in a valley (a relative minimum). Unfortunately, if the sign doesn't change, the critical point is a mere flatland, neither a maximum nor a minimum. Our function doesn't have a constant pattern of derivative signs around the critical points, indicating true changes in the terrain—that is, genuine relative extrema.

Second Derivative Test

The Second Derivative Test further scrutinizes our critical points to determine if we're standing on a peak or dipping into a valley. By taking the derivative of the first derivative, we get the second derivative, f''(x) = 6x. Now, we probe into the nature of the critical points.
For x = 1, where f''(1) = 6 which is greater than 0, it reveals a concave up parabola, a sure sign of a valley or a relative minimum. On the flip side, for x = -1, where f''(-1) = -6 less than 0, shows a concave down parabola, synonymous with a peak or relative maximum. This test is a decisive method to confirm the type of extremum at each critical point without much fuss.