Question

For each of the following functions of one variable, determine all local maxima and minima and indicate points of inflection (where \(/^{\prime \prime}=0\) ): a. \(f(x)=4 x^{3}-12 x\) b. \(f(x)=4 x-x^{2}\) c. \(f(x)=x^{3}\)

Short answer

Question: For the following functions, find the local maxima, local minima, and points of inflection: a. \(f(x)=4x^3-12x\) b. \(f(x)=4x-x^2\) c. \(f(x)=x^3\) Answer: a. Local maximum at \(x=-1\), local minimum at \(x=1\), and no inflection points. b. Local maximum at \(x=2\), no local minimum, and no inflection points. c. No local maxima or minima, and inflection point at \(x=0\).

Step-by-step solution

a. \(f(x)=4x^3-12x\)

First, we are going to find the first derivative of the function using the power rule: \(f'(x)=3\cdot4x^{3-1}-1\cdot12x^{1-1}=12x^2-12\) Now, let's set the first derivative equal to 0 to determine the critical points (i.e., where the local maxima and minima could occur): \(12x^2-12=0 \Rightarrow x^2=1 \Rightarrow x=-1,1\) Next, we need to find the second derivative of the function to determine the concavity of the function: \(f''(x)=2\cdot12x^{2-1}=24x\) Now, let's evaluate the second derivative at the critical points to determine if these points are indeed local maxima, local minima, or inflection points: - For \(x=-1\), we have \(f''(-1)=-24\). Since the second derivative is negative, this indicates a local maximum at \(x=-1\). - For \(x=1\), we have \(f''(1)=24\). Since the second derivative is positive, this indicates a local minimum at \(x=1\). Thus, there is a local maximum at \(x=-1\) and a local minimum at \(x=1\). There are no inflection points as the second derivative is never equal to 0.

b. \(f(x)=4x-x^2\)

First, we need to find the first derivative of the function: \(f'(x)=4-2x\) Now, let's set the first derivative equal to 0 to determine the critical points: \(4-2x=0 \Rightarrow x=2\) Next, we need to find the second derivative of the function to determine the concavity of the function: \(f''(x)=-2\) Since the second derivative is negative, this indicates a local maximum at \(x=2\). There are no inflection points, as the second derivative is always negative and never equals 0.

c. \(f(x)=x^3\)

First, let's find the first derivative of the function: \(f'(x)=3x^2\) Now, let's set the first derivative equal to 0 to determine the critical points: \(3x^2=0 \Rightarrow x=0\) Next, we need to find the second derivative of the function to determine the concavity of the function: \(f''(x)=6x\) Now, let's evaluate the second derivative at the critical point to determine if this point is a local maximum, local minimum, or inflection point: - For \(x=0\), we have \(f''(0)=0\). Since the second derivative is equal to 0, there is an inflection point at \(x=0\). Thus, there are no local maxima or minima, and there is an inflection point at \(x=0\).

Key concepts

Local Maxima and Minima

Understanding local maxima and minima is a crucial part of microeconomic theory, which emphasizes decision-making and efficiency.

Let's consider a function that represents some economic quantity — for example, profit as a function of the quantity of goods produced. A local maximum of this function is a point where the profit is greater than at any nearby points, signifying the highest achievable profit in a local neighborhood around that point. Conversely, a local minimum is where the profit is less than at any adjacent point, indicating the lowest profits nearby.

Mathematically, a function's derivative provides information about its slope. At local maxima and minima, the slope is zero because the function levels off before changing direction. To find these points, one would calculate the first derivative of the function and set it equal to zero to find potential candidates, known as critical points.

Inflection Points

Moving to the concept of inflection points, these are crucial for economic models as they signify changes in the direction of curvature on a graph of a function. An inflection point occurs where a function shifts from concave upward (like a cup) to concave downward (like a cap), or vice versa.

An easy analogy to understand inflection points is a car driving over a hill; the hilltop would be a local maximum, and as the car drives down and the road begins to level out, the point where it stops being concave and starts becoming convex is an inflection point.

To determine inflection points mathematically, we investigate where the second derivative of the function changes sign. The second derivative gives information about the concavity of a function. If the second derivative is equal to zero and changes sign at that point, we have an inflection point. For the function given in the exercise, only function c, with the form f(x)=x^3, has an inflection point at x=0, because the second derivative at this point is zero, indicating a shift in concavity.

Second Derivative Test

The second derivative test is a practical tool for categorizing critical points as local maxima or minima. After identifying critical points by setting the first derivative of a function to zero, the second derivative test can determine the nature of these points.

According to the second derivative test, if the second derivative at a critical point is positive, the function is concave upward at that point, indicating a local minimum. If it's negative, the function is concave downward, and we have a local maximum. For example, from the solutions given, we can see function a, f(x)=4x^3-12x, assessed at critical points x=-1 and x=1. Here, the second derivative test tells us that x=-1 is a local maximum while x=1 is a local minimum.

However, when the second derivative is zero, the test is inconclusive, and we may resort to other methods to evaluate the critical point. This test helps in determining not only the nature of the extremities but also the general 'shape' of the graph around these points.