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Chapter 4: Problem 2

What are local maximum and minimum values of a function?

Short Answer

Expert verified
Answer: To find local maximum and minimum values of a function, follow these steps: 1. Understand the terms local maximum and local minimum: Local maximum is a point where the function has a greater value than nearby points, while local minimum is a point where the function has a lower value than nearby points. 2. Identify critical points: Find the derivative of the function, set the derivative equal to zero, and solve for the x-values at which this occurs. Also check for any points where the derivative is undefined. 3. Determine if the critical points correspond to local maximum or minimum values: Use the second derivative test by finding the second derivative, plugging in the x-values of the critical points, and analyzing the results. If the second derivative is positive, the critical point is a local minimum; if negative, it is a local maximum. If the test is inconclusive, use a different method such as analyzing the signs of the first derivative around the critical point.

Step by step solution

01

Understand the terms local maximum and local minimum

A local maximum of a function is a point where the function has a greater value than all nearby points, while a local minimum is a point where the function has a lower value than all nearby points. These are also referred to as relative maximum and minimum values, respectively.
02

Identify critical points

Critical points are points on the function's graph where the derivative is either zero or does not exist. To find the critical points, first find the derivative of the function. Next, set the derivative equal to zero and solve for the x-values at which this occurs. Also check for any points where the derivative is undefined (for example, if the derivative is a fraction and the denominator is zero).
03

Determine if the critical points correspond to local maximum or minimum values

To check if a critical point is a local maximum or minimum, we can use the second derivative test. First, find the second derivative of the function. Then, plug the x-values of the critical points into the second derivative. If the value of the second derivative is positive, the critical point corresponds to a local minimum, and if it is negative, the critical point corresponds to a local maximum. If the second derivative is zero or undefined at a critical point, the test is inconclusive, and you would need to use a different method, such as analyzing the signs of the first derivative around the critical point. Using these steps, you can identify the local maximum and minimum values of a function.

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Key Concepts

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

Critical Points
In calculus, critical points play a crucial role when identifying local maxima and minima of a function. Simply put, these are points on the graph where the function's derivative is zero or undefined. Why do they matter? Because at these specific spots, the function might change its direction鈥攅ither peaking at a high point (maximum) or dipping to a low point (minimum).

Here鈥檚 how to find them:
  • Calculate the derivative of your function.
  • Set this derivative to zero and solve for the variable, usually represented as 'x'. These solutions are potential critical points.
  • Check if the derivative is undefined at any points. Functions with denominators that can equal zero are particularly tricky and need your attention.
Critical points are just potential starting points for finding local extrema. The next step is determining the nature of these points, which brings us to examining derivatives further.
Derivative
The derivative of a function is foundational in calculus as it provides information about the rate of change at any given point on the function itself. Imagine tracing the curve of a roller coaster. The steepness of each section can be described by the derivative, showing you whether the function is increasing or decreasing.

To find a function's derivative:
  • Differential calculus is used to create a derivative, capturing the slope of the tangent line at each point.
  • The notation used is often expressed as \(f'(x)\) or \(dy/dx\), signifying the derivative of the function \(f(x)\).
Having the derivative lets you find critical points, as described previously. More than that, it tells you about the general behavior of the function: is it leaning upwards, or is it diving down? Understanding this behavior helps in interpreting critical points accurately as possible maximums, minimums, or neither.
Second Derivative Test
Once you've found the critical points, the second derivative test helps to analyze whether these are points of local maximum or minimum. This test involves evaluating the second derivative of the function at these critical points.

The steps include:
  • Find the second derivative, \(f''(x)\), which provides the concavity of the function. "Concavity" refers to whether the function curves upwards or downwards at a certain point.
  • Plug the x-values of the critical points into the second derivative.
  • If \(f''(x) > 0\), the function is concave up, indicating a local minimum at that critical point.
  • If \(f''(x) < 0\), the function is concave down, signifying a local maximum.
  • A result of \(f''(x) = 0\) means the test is inconclusive, and you may need to check using another method or examining the first derivative's sign changes near the point to decide on its nature.
The second derivative test is a handy tool, simplifying what could be a complicated analysis by giving straightforward criteria to classify critical points.

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Most popular questions from this chapter

Use analytical methods to evaluate the following limits. $$\lim _{x \rightarrow \infty}\left(x^{2} e^{1 / x}-x^{2}-x\right)$$

Economists use demand functions to describe how much of a commodity can be sold at varying prices. For example, the demand function \(D(p)=500-10 p\) says that at a price of \(p=10,\) a quantity of \(D(10)=400\) units of the commodity can be sold. The elasticity \(E=\frac{d D}{d p} \frac{p}{D}\) of the demand gives the approximate percent change in the demand for every \(1 \%\) change in the price. a. Compute the elasticity of the demand function \(D(p)=500-10 p\) b. If the price is \(\$ 12\) and increases by \(4.5 \%,\) what is the approximate percent change in the demand? c. Show that for the linear demand function \(D(p)=a-b p\) where \(a\) and \(b\) are positive real numbers, the elasticity is a decreasing function, for \(p \geq 0\) and \(p \neq a / b\) d. Show that the demand function \(D(p)=a / p^{b}\), where \(a\) and \(b\) are positive real numbers, has a constant elasticity for all positive prices.

Show that the general quartic (fourth-degree) polynomial \(f(x)=x^{4}+a x^{3}+b x^{2}+c x+d\) has either zero or two inflection points, and the latter case occurs provided that \(b<3 a^{2} / 8.\)

Linear approximation a. Write an equation of the line that represents the linear approximation to the following functions at a. b. Graph the function and the linear approximation at a. c. Use the linear approximation to estimate the given quantity. d. Compute the percent error in your approximation. $$f(x)=\tan x ; a=0 ; \tan 3^{\circ}$$

Evaluate the following limits in terms of the parameters a and b, which are positive real numbers. In each case, graph the function for specific values of the parameters to check your results. $$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}$$
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