We've already looked at this—although we had to do it with a calculator at the time. Now we can do it by hand, with Calculus!
Recall that relative extrema are extremes on some open interval. Absolute extrema are extremes for the whole graph (all real numbers).
I've said previously (in class) that maximums and minimums must have horizontal tangent lines. Now we're going to use that.
A critical number is any value of x where f ′(x) = 0, or f ′(x) fails to exist. f ′(x) will fail to exist at any values of x where the function has a sharp point (called a cusp). As an example, the function g(x) = |x| has a cusp at x = 0.
The only time we get a maximum or minimum at a location other than a critical value is when we restrict our attention to a closed interval. In that case, you must also check the function values at the ends of the interval.
Examples:
[1.] Find any critical numbers for .
Well, you'll need the derivative: . The only way that a fraction can equal zero is for the numerator to equal zero, so we get x(x + 2) = 0. So we have critical numbers at x = 0 and x = -2. Since f(x) is a rational function, it is continuous and differentiable on its domain—which excludes x = -1. Thus, the critical numbers are -1, 0 and 2.
[2.] Locate any relative extrema for . Find the extreme values.
. This is defined for all reals, so any critical numbers come from its zeros. ⇒ x = 0 or x = 1. At x = 0, y = 0; and at x = 1, y = 1. Now, add the endpoints of the given interval, and we obtain the list of candidate locations for absolute extrema: -1, 0 and 1. The function values at these points are (respectively) 5, 0 and 1. Thus, the absolute maximum is at x = -1, and the absolute minimum is at x = 0.
Again, this topic is not new—we're just going to look at it again, in light of The Calculus.
Increasing means that the function is going up—what does that mean about its derivative?
If f ′(x) > 0 on some interval (a, b), then that means that the function is increasing on that interval.
If f ′(x) < 0 on some interval (a, b), then that means that the function is decreasing on that interval.
Examples:
[3.] Determine intervals where f(x) = 2x³ - 3x² + 3 is increasing and decreasing.
First, take the derivative: f(x) = 2x³ - 3x² + 3 ⇒ f ′(x) = 6x² - 6x. Set this equal to zero to find the critical numbers: 6x² - 6x = 0 ⇒ x² - x = 0 ⇒ x(x - 1) = 0 ⇒ x = 0 or x = 1. For x < 0, the derivative is positive; for x between 0 and 1, the derivative is negative; for x > 0, the derivative is positive. So f(x) is increasing on the intervals (-∞, 0) and (1, ∞); f(x) is decreasing on the interval (0, 1).
[4.] Determine intervals where is increasing and decreasing.
The derivative: . This has only one critical value (at x = 1), but even that doesn't matter much—for any valid value of x, the derivative is negative (since the denominator can only be positive)! Thus, the function is decreasing on the intervals (-∞, 1) and (1, ∞).
Now, remember that a maximum occurs between increasing and decreasing—in other words, when the function switches from increasing to decreasing, then there may be a maximum point in there. Similarly, a minimum occurs between decreasing and increasing. So, we can find maximums and minimums by looking for sign changes in the first derivative! Don't forget that the function must be defined at the value in question…
Examples:
[5.] The function f(x) = 2x³ - 3x² + 3 has a maximum at x = 0 (since that is where the derivative switches from + to -), and a minimum at x = 1.
[6.] Locate any relative extrema for the function .
First of all, perhaps we should note that this function is only defined if 4 - x² ≥ 0, which solves to -2 ≤ x ≤ 2.
Well—how about the derivative? . Now, set that equal to zero. Note that the middle part has a negative exponent, and thus cannot make the whole thing become zero—just the (- 2x) has that power. (- 2x) = 0 ⇒ x = 0. For values of x less than 0, the derivative will be positive (the middle part is always positive, because of the radical). For values of x greater than 0, the derivative will be negative. Since the derivative changes from positive to negative, there is a maximum at x = 0.
On an interval where f ′(x) is increasing, f(x) is concave up.
On an interval where f ′(x) is decreasing, f(x) is concave down.
Now, we know the relationship between intervals where f(x) is increasing/decreasing, and f ′(x) is positive/negative—with concavity, we're talking about the derivative increasing/decreasing…so when the derivative is increasing/decreasing, f ″(x) will be positive/negative.
So:
On an interval where f ″(x) > 0, the graph of f(x) will be concave up.
On an interval where f ″(x) < 0, the graph of f(x) will be concave down.
When the first derivative, f ′(x), changes between positive and negative, there is a critical number (and maybe an extremum).
When the second derivative, f ″(x), changes between positive and negative, there is a point of inflection (provided there is a tangent line at that point). Note, however, that f ″(x) = 0 does not guarantee the existence of a point of inflection.
If f ′(c) = 0 and f ″(c) > 0, then there is a relative minimum at x = c.
If f ′(c) = 0 and f ″(c) < 0, then there is a relative maximum at x = c.
Examples:
[7.] Locate any points of inflection on the graph of y = x³ - 6x² - 135x.
Derivative time! y′ = 3x² - 12x - 135, and y″ = 6x - 12. 6x - 12 = 0 ⇒ x = 2. There is one point of inflection at x = 2.
[8.] Use the second derivative test to locate any extrema on the graph of .
First Derivative: y′ = 12x³ - 12x². Critical numbers: 12x³ - 12x² = 0 ⇒ 12x²(x - 1) = 0 ⇒ x = 0 or x = 1.
Second Derivative: y″ = 36x² - 24x. When x = 1, y″ > 0, which means that we have a minimum at x = 1.
When x = 0, y″ = 0. This doesn't give us any information; we'll need to look at a sign of y′ on either side of x = 0. For values of x slightly less than 0, y′ is negative. For values of x slightly greater than 0, y′ is negative. Thus, there is no extremum at x = 0.
The only extremum is a minimum at (1, -1).
Back in the days before graphing calculators, this was much more important…now it simply serves as conclusive proof of the number/location of extrema.
If you're going to sketch a curve, then find the domain; all intercepts (x and y); critical numbers (and extrema); points of inflection. Plot what you know, and connect the points with a smooth curve.
Examples:
[9.] Sketch . Since this is a polynomial, the domain is all real numbers. To find the x-intercepts: ⇒ ⇒ x = 0 or . The y-intercept is at (0, 0). The only extremum is at (1, -1). As for the points of inflection: 36x² - 24x = 0 ⇒ 3x² - 2x = 0 ⇒ x(3x - 2) = 0 ⇒ x = 0 or x = . Plot (I've marked the points of inflection as open dots):
…and connect!
[10.] Sketch y = x³ - 3x² + 2.
Another polynomial—no domain problems.
x-intercepts: x³ - 3x² + 2 = 0. Notice that x = 1 works: (x - 1)(x² - 2x - 2) = 0. The quadratic formula helps for the other two intercepts: .
y-intercept: (0, 2).
y′ = 3x² - 6x. 3x² - 6x = 0 ⇒ x(x - 2) = 0 ⇒ critical numbers at x = 0 and x = 2.
y″ = 6x - 6. At x = 0, this is negative, indicating a maximum: (0, 2). At x = 2, this is positive, indicating a minimum: (2, -2). 6x - 6 = 0 ⇒ x = 1. One point of inflection at x = 1: (1, 0).
Plot:
Connect:
Word problems!
If you're trying to find a maximum or minimum of something, then you need an equation for that something! Use what we've learned to locate the extrema.
Examples:
[11.] An open rectangular box is to be made from a piece of cardboard 8 in wide and 15 in long by cutting squares from the corners and folding up the resulting flaps. Find the dimensions of the box that produce a maximum volume.
First of all, the volume equation: . Now, since we want to find a maximum volume, we need to take the derivative and locate any critical numbers (so as to locate any extrema). First, perhaps we should rewrite the volume equation: . So . Set this equal to zero: . Fortunately, this factors! (3x - 5)(x - 6) = 0 ⇒ x = 6 or .
Let's check the second derivative to see which one is a maximum: . Letting x = 6 makes V″ positive, meaning that we have a minimum there. Letting makes V″ negative, meaning that we have a maximum (note that x = 6 isn't even in the domain of the problem—that would cut more cardboard than there is!).
So the height is ; the length is ; and the width is .
[12.] Find two positive numbers whose sum is 36 and whose product is as large as possible.
Well, we could let the two numbers be x and y, which would produce the equation x + y = 36, and force us to maximize the product xy…but then we don't have a function for the product, since it uses two variables! Do we really need two?
No! Since x + y = 36, the second number can just be 36 - x. Then we need to maximize the product x(36 - x), which is a function.
, so . This has only one critical number, at x = 18. Second derivative time! P″(x) = -2, which tells us that at x = 18, there is a maximum (since the product function is concave down). Thus, the two numbers are 18 and 18.
Page last validated 2010-08-15