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Chapter 15: Problem 6

Explain how the Second Derivative Test is used.

Short Answer

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Answer: The Second Derivative Test is a method to determine the nature of critical points (local maxima, local minima, or saddle points) of a function. To apply the test: 1. Find the first derivative (f'(x)) and locate the critical points by setting f'(x) equal to 0 or finding points where f'(x) is undefined. 2. Find the second derivative (f''(x)). 3. Evaluate f''(x) for each critical point. - If f''(x) > 0, the critical point is a local minimum (concave up). - If f''(x) < 0, the critical point is a local maximum (concave down). - If f''(x) = 0 or undefined, the test is inconclusive, and another method like the First Derivative Test should be used to determine the critical point's nature.

Step by step solution

01

Understand Critical Points

A critical point of a function is any point on the graph of the function where the first derivative is either equal to zero or does not exist. These points are important because they can be potential local maximums, local minimums, or saddle points.
02

Find the Critical Points

To find the critical points of a given function, first, find the first derivative of the function. Then, set the first derivative equal to zero or find points where the derivative is undefined to find the critical points.
03

Understand the Second Derivative Test

The Second Derivative Test is a method to determine the nature of the critical points (local maxima, local minima, or saddle points). It involves evaluating the second derivative of the function at the critical points. The second derivative provides information about the concavity of the graph, which can help determine the type of critical point.
04

Apply the Second Derivative Test

Given a function f(x), apply the following steps of the Second Derivative Test: 1. Find the first derivative, f'(x), and locate the critical points by finding the values of x for which f'(x) = 0 or is undefined. 2. Find the second derivative, f''(x). 3. For each critical point, evaluate the second derivative, f''(x), at that point. The test outcomes can be interpreted as follows: - If f''(x) > 0 at a critical point, the function is concave up at that point, and the critical point is a local minimum. - If f''(x) < 0 at a critical point, the function is concave down at that point, and the critical point is a local maximum. - If f''(x) = 0 or is undefined at a critical point, the test is inconclusive, and one should resort to another method, such as the First Derivative Test, to determine the nature of the critical point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Critical Points
Imagine driving along a hilly road; critical points are akin to the very top of a hill or the very bottom of a valley where your car would come to a temporary halt before ascending or descending. In mathematical terms, critical points occur where the slope of a function's graph鈥攐r its first derivative鈥攊s zero or undefined. These points are pivotal in analyzing functions because they potentially signify where the function reaches its highest or lowest value in a local area, known as local maxima and minima, or, less frequently, saddle points, which are neither.
Concavity and the Second Derivative
The concavity of a graph tells us the direction in which it curves. A function is concave up if it curves upwards, resembling a happy smile. Conversely, a function is concave down if it looks like a frown. The second derivative of a function, when evaluated at a particular point, informs us about the graph's concavity at that point. If the second derivative is positive, the graph is concave up. If negative, it's concave down. This attribute is crucial because it helps determine the nature of the critical points unearthed in the function's graph; whether they're positioned at the summit of a hill (local maxima) or in the depth of a valley (local minima).
Local Maxima and Minima
Local maxima and minima represent the peaks and troughs found in various segments of a function's graph. A local maximum is the highest point within a particular range, functioning as a ceiling that the function doesn't breach. Conversely, a local minimum is like the floorboard of a small region in the graph, being the lowest point the function reaches before it starts climbing up again. The second derivative test is especially handy when identifying if a critical point is a local maxima or minima. If the function curves up and forms a valley at the critical point, it's a local minima, but if it curves down and forms a hilltop, you've found a local maxima. These local extreme values are vital in various applications, such as optimization problems where one needs to find the best solution, be it a maximum profit or a minimum cost.

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Most popular questions from this chapter

Problems with two constraints Given a differentiable function \(w=f(x, y, z),\) the goal is to find its absolute maximum and minimum values (assuming they exist) subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0,\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0,\) and \(h(x, y, z)=0\)

Find the values of \(K\) and \(L\) that maximize the following production functions subject to the given constraint, assuming \(K \geq 0\) and \(L \geq 0\) $$P=f(K, L)=10 K^{1 / 3} L^{2 / 3} \text { for } 30 K+60 L=360$$

Graph several level curves of the following functions using the given window. Label at least two level curves with their z-values. $$z=x-y^{2} ;[0,4] \times[-2,2]$$

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) $$ \text { b. } f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}} $$ This property has the following interpretation. Suppose a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum. (Source: Mathematics Magazine, May \(1985,\) and Calculus and Analytical Geometry, 2 nd ed., Philip Gillett, 1984 )

Find the domain of the following functions. If possible, give a description of the domains (for example, all points outside a sphere of radius 1 centered at the origin). $$Q(x, y, z)=\frac{10}{1+x^{2}+y^{2}+4 z^{2}}$$
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