f (x) has an absolute minimum value f (c) at the point x = c in its domain if

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1 Definitions - Absolute maximum and minimum values f (x) has an absolute maximum value f (c) at the point x = c in its domain if f (x) f (c) holds for every x in the domain of f (x). f (x) has an absolute minimum value f (c) at the point x = c in its domain if f (x) f (c) holds for every x in the domain of f (x). The Extreme Value Theorem (EVT) If f (x) is continuous on the closed interval [a, b], then: f (x) MUST attain an absolute maximum on [a, b], and f (x) MUST attain an absolute minimum on [a, b]. Definitions - Local maximum and minimum values f (x) has a local maximum value f (c) at the point x = c in its domain if f (x) f (c) holds for every x in an interval around c. f (x) has a local minimum value f (c) at the point x = c in its domain if f (x) f (c) holds for every x in an interval around c. Note: Some textbooks require the interval to be open in this definition (ours does not). AMAT 217 (University of Calgary) Fall / 20

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3 Definitions - Special Points c is a critical point of f (x) if c is in the domain of f and f (c) = 0. c is a singular point of f (x) if c is in the domain of f and f (c) is undefined. c is an end point of the domain of f (x) if c is in the domain of f and c does not belong to any open interval contained in the domain of f. Fermat s Theorem (FT) Let f (x) be defined on an interval I. If f (x) has a local maximum or minimum at x = c, then either: x = c is a critical point of f (x), or x = c is a singular point of f (x), or x = c is a endpoint of the interval I. Test for absolute extrema of continuous functions on closed intervals: 1 Verify that the function is continuous on the interval [a, b]. 2 Find all special points of f (x) that are in the interval [a, b]. 3 Evaluate the function at the special points from step 2 (including the end points). 4 Identify the absolute extrema by comparing values. AMAT 217 (University of Calgary) Fall / 20

4 Example Find the absolute extrema of f (x) = 1 3 x x 2 2x + 1 on the interval [ 3, 5]. Step 1: The function is a polynomial so is continuous on [ 3, 5] Step 2: The derivative is: f (x) = x 2 + x 2 = (x + 2)(x 1) Both x = 1 and x = 2 are critical points (there are no singular points) Step 3: Plugging in endpoints and critical points belonging to [ 3, 5] gives: Step 4: Comparing gives: Critical Points: f ( 2) = f (1) = Singular Points: None Endpoints: f ( 3) = 2.5 f (5) = The absolute maximum is the largest the function can be on [ 3, 5] The absolute maximum of f (x) is which occurs at x = 5 The absolute minimum is the smallest the function can be on [ 3, 5] The absolute minimum is which occurs at x = 1 AMAT 217 (University of Calgary) Fall / 20

5 Curve Sketching We next look at an application of derivatives. Goal: What does f (x) = x 3 + 6x 2 15x + 7 look like? AMAT 217 (University of Calgary) Fall / 20

6 The Increasing and Decreasing Test 1 If f (x) > 0 on some interval I, then f (x) is increasing on I. 2 If f (x) < 0 on some interval I, then f (x) is decreasing on I. Example Determine all intervals where f (x) = x 3 + 6x 2 15x + 7 is increasing or decreasing. Solution: We will need the derivative: f (x) = 3x x 15 = 3(x 2 + 4x 5) = 3(x 1)(x + 5). We draw a number line and put the special points on the number line: The special points split the number line up into subintervals. Choose test points and sub into f (x) to determine inc/dec intervals. Decreasing on the interval ( 5, 1) and increasing on the intervals (, 5) (1, ). Caution: Sub test points into the derivative, not the original function. AMAT 217 (University of Calgary) Fall / 20

7 The First Derivative Test Suppose that x = c is a critical point or singular point of f (x). If f (x) is increasing (f (x) > 0) immediately to the left of x = c and decreasing (f (x) < 0) immediately to the right of x = c then x = c is a local maximum. If f (x) is decreasing (f (x) < 0) immediately to the left of x = c and increasing (f (x) > 0) immediately to the right of x = c then x = c is a local minimum. AMAT 217 (University of Calgary) Fall / 20

8 Example Find and classify all the critical points of the following function: f (x) = x 3 + 6x 2 15x + 7. Solution: From before, the derivative is: f (x) = 3(x 1)(x + 5), with critical points x = 5, 1 and intervals of increasing/decreasing as illustrated: Therefore, x = 1 is a local minimum and x = 5 is a local maximum. AMAT 217 (University of Calgary) Fall / 20

9 Concavity Concavity is best illustrated with a picture: Informally, a function is concave up if it opens up. A function is concave down if it opens down. Concavity has to do with the shape of the graph not whether the graph increases or decreases. Often it helps to think of the four possible combinations of concavity with increasing/decreasing: AMAT 217 (University of Calgary) Fall / 20

10 Mathematical definition of concavity: Given the function f (x) and an interval I, then f (x) is concave up on I if all the tangents to the curve on I are below the graph of f (x) f (x) is concave down on I if all the tangents to the curve on I are above the graph of f (x) Definition: A point x = c is called an inflection point if the function is continuous at the point and the concavity of the graph changes at that point. Question: What do you notice about the slopes of the tangent lines? AMAT 217 (University of Calgary) Fall / 20

11 Concavity Test If f (x) > 0 on some interval I, then f (x) is concave up on I. If f (x) < 0 on some interval I, then f (x) is concave down on I. Possible inflection points are where f (x) = 0 or DNE (check the concavity on both sides). Example Identify the intervals of concavity and inflection points for f (x) = x 3 + 6x 2 15x + 7. The process for determining concavity is almost identical to that of determining increasing/decreasing (except we use the second derivative instead). The second derivative is: f (x) = 6(x + 2). We first determine possible inflection points (where f (x) = 0 or f (x) does not exist). In our case, f (x) = 0 when x = 2. Draw a number line and sub test points into f (x): f (x) is concave down on (, 2) and concave up on ( 2, ). Since concavity changes at x = 2, then x = 2 is an inflection point. AMAT 217 (University of Calgary) Fall / 20

12 Goal: What does f (x) = x 3 + 6x 2 15x + 7 look like? x = 1 is a local minimum x = 5 is a local maximum x = 2 is an inflection point decreasing on the interval ( 5, 1) increasing on the intervals (, 5) (1, ) concave down on the interval (, 2) concave up on the interval ( 2, ) AMAT 217 (University of Calgary) Fall / 20

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14 Second Derivative Test We can also use second derivative to classify critical points. Second Derivative Test Suppose that x = c is a critical number of f (x) such that f (c) = 0 and that f (x) is continuous in a region around x = c. Then, If f (c) < 0, then x = c is a local maximum. If f (c) > 0, then x = c is a local minimum. If f (c) = 0, then the second derivative test has no conclusion about c. AMAT 217 (University of Calgary) Fall / 20

15 Example Classify the critical points of using the second derivative test. f (x) = x 3 + 6x 2 15x + 7 Solution: Earlier we saw the critical points to be x = 1 and x = 5. We also noted that f (x) = 6(x + 2). Thus, f (1) = 18 > 0 and f ( 5) = 18 < 0. Therefore, x = 1 is a local minimum and x = 5 is a local maximum. AMAT 217 (University of Calgary) Fall / 20

16 Guidelines for Curve Sketching: Gather as much information as you can on f (x), including: Curve sketching guidelines Domain of f. x and y intercepts. Symmetry: Is it an even or odd function? Periodic? Horizontal, vertical and slant asymptotes. Intervals of increase/decrease. Find critical points and classify as either local max, local min or neither. Concavity and inflection points. If possible, evaluate the function at a few key points (critical and inflection points). Sketch the graph. AMAT 217 (University of Calgary) Fall / 20

17 Example Sketch the graph of y = f (x) where f (x) = 2x 2 x 2 1. Solution: Domain: The domain is {x : x 2 1 0} = {x : x ±1} = (, 1) ( 1, 1) (1, ) Intercepts: There is an x-intercept at x = 0. The y intercept is y = 0. Symmetry: f ( x) = f (x), so f is an even function (symmetric about y-axis) 2x 2 Asymptotes: lim x ± x 2 1 = lim 2 x ± 1 1/x 2 = 2. Therefore, y = 2 is a horizontal asymptote. Now the denominator is 0 at x = ±1, so we compute: 2x 2 lim x 1 + x 2 1 = +, lim 2x 2 x 1 x 2 1 =, lim 2x 2 x 1 + x 2 1 =, So the lines x = 1 and x = 1 are vertical asymptotes. lim 2x 2 x 1 x 2 1 = +. AMAT 217 (University of Calgary) Fall / 20

18 Example Sketch the graph of y = f (x) where f (x) = 2x 2 x 2 1. Solution (continued): Critical points/inc/dec: For critical points we take the derivative: f (x) = 4x(x 2 1) 2x 2 2x (x 2 1) 2 = 4x (x 2 1) 2. Note that f (x) = 0 when x = 0 (the top is zero). Also, f (x) = DNE when x = ±1 (the bottom is zero). As x = ±1 is not in the domain of f (x), the only critical point is x = 0 (recall that to be a critical point we need it to be in the domain of the original function). Drawing a number line and including all of the split points of f (x) we have: Thus f is increasing on (, 1) ( 1, 0) and decreasing on (0, 1) (1, ). By the first derivative test, x = 0 is a local max. AMAT 217 (University of Calgary) Fall / 20

19 Example Sketch the graph of y = f (x) where f (x) = 2x 2 x 2 1. Solution (continued): Concavity/inf points: For possible inflection points we take the second derivative: f (x) = 12x (x 2 1) 3 The top is never zero. Also, the bottom is only zero when x = ±1 (but these are not in domain). Thus, there are no possible inflection points to consider. Drawing a number line and including all of the split points of f (x) we have: Hence f is concave up on (, 1) (1, ), concave down on ( 1, 1). AMAT 217 (University of Calgary) Fall / 20

20 Combine on a single number line: Note that there is a horizontal asymptote at y = 2 and that the curve has x-int of x = 0 and y-int of y = 0. Therefore, a sketch of f (x) is as follows: AMAT 217 (University of Calgary) Fall / 20