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Applications of the Second Derivative BY DR. JULIA ARNOLD USING TAN’S 5TH EDITION APPLIED CALCULUS FOR THE MANAGERIAL , LIFE, AND SOCIAL SCIENCES TEXT I recommend that you view the power point as a slide presentation first. After viewing then print the slides as notes according to the instructions below. Under view, click on slide show. To print the slides as notes go to the file command and select print. On the window that comes up toward the bottom it may say print what and next to that is a drop down menu. Select handouts and the number you want per page. Determining the Intervals of Concavity A function is concave up on an interval (a,b) if f’ is increasing on (a,b) Slope is -.5 Slope is .5 Since the slopes are Slope is 0 -.5,0,.5 and since these numbers are increasing the function is concave up. Associate concave up with a bowl shape and it is right side up so everything is stays in the bowl. Hence “positive” + A function is concave down on an interval (a,b) if f’(x) is decreasing on (a,b). Slope is 0 Slope is 2 Since the slopes are 2,0,-2 and since these numbers are decreasing the function is concave down. Slope is -2 Associate concave down with a bowl shape and it is upside down so everything is falling out of the bowl. Hence “negative” - Since the first derivative can tell us where a function is increasing and where a function is decreasing, then if that function is f’, the second derivative will tell us where f’(x) is increasing or decreasing. Concave Down First Derivative is decreasing. Second Derivative is negative. Since the first derivative can tell us where a function is increasing or decreasing, then if that function is f’, the second derivative will tell us where f’(x) is increasing or decreasing. The function is concave up. The first derivative is increasing. The second derivative is positive. The Rule: 1. If f”(x) > 0 for each value of x in (a,b) then f is concave upward on (a,b). 2. If f”(x) < 0 for each value of x in (a,b) then f is concave downward on (a,b). Example 1: f(x)= x2 is a function which is only concave up on the entire x axis and f(x)= -x2 is concave down on the entire x axis. Concave up Concave down f’(x)= 2x f”(x)= 2 which is always a positive number f’(x)= -2x f”(x)= -2 which is always a negative number Example 1: Determine the intervals where the function is concave up and concave down f(x) x3 3x2 24x 32 Step 1: Find the second derivative of f f(x) x3 3x2 24x 32 f(x) 3x2 6x 24 f(x) 6x 6 Step 2: Set the second derivative equal to 0 and solve for x. f(x) 6x 6 0 6x 6 6 6x 1x 1 Draw a number line and insert 1 on it. Two intervals are formed. ,1 1, Test Points ,1 0 2 1, 1 Substitute in f”(x)=6x-6 Resulting Sign -6 6 + Interpretation of resulting signs is that the function is concave down on ,1 and concave up on 1, . The red function is concave down, then when it reaches x = 1 it switches concavity to upward. 1 Example 2: Find the intervals where the function f(x) x x is concave up and concave down. Remember that the domain does not contain 0. Thus the intervals over which the function is defined are ,00, Next, find f’(x). 1 x 1 f(x ) 1 2 x f( x ) x Now find f”(x) f(x) 1 1 2 1 x x2 f(x) 2x 3 f(x) 2 x3 Since f”(x) 0, we look at where it is undefined or discontinuous, which would be at x = 0. ,0 Thus the only intervals are: 0, Intervals Test Point Substitution into f(x) ,0 0, -1 1 2/-1 = -2 2/1 =2 2 x3 Result + Interpretation of results: On 0, f”(x) > 0 f(x) is concave up. On ,0 , since f”(x) < 0, f(x) is concave down. Example 3: Find the intervals where the function f(x) x is concave up and concave down. Find f’(x). Now find f”(x) 1 f(x) x 2 2 f(x) x 3 3 3 3 x 2 3 2 f(x) x 3 1 3 2 3 3 x 4 2 3 2 f (x ) x 4 9 9x 3 Since f” can not be 0, we choose x = 0 where f”(x) is undefined or discontinuous. Thus the intervals under consideration are: ,0 0, 2 3 Choose a test point: Test Point ,0 0, Substitution into f’(x) -1 1 f(1) 2 4 3 9(1) 2 2 f(1) 4 9 913 2 9 Result - Interpretation of results: ,0 0, On this interval, f”(x) < 0 so f(x) is concave down. On this interval, f”(x) < 0 so f(x) is concave down. f(x) is concave down f(x) is concave down The Rule: 1.Find all values of x for which f”(x)=0 or f” is discontinuous and identify the open intervals determined by these points. 2.Select a test point c in each interval found in step 1 and determine the sign of f”(c) in that interval. A. If f”(c) >0 in that interval, f is concave upward on that interval. B. If f”(c) <0 in that interval, f is concave downward on that interval. Practice problem: Find the interval where the function is concave up or concave down. g(x) x 2 The Rule: 1.Find all values of x for which f”(x)=0 or f” is discontinuous and identify the open intervals determined by these points. 2.Select a test point c in each interval found in step 1 and determine the sign of f”(c) in that interval. A. If f”(c) >0 in that interval, f is concave upward on that interval. B. If f”(c) <0 in that interval, f is concave downward on that interval. Which is the correct first step? g( x ) x 2 g(x) x 2 1 g( x ) x 2 2 1 1 2 1 2 1 1 x 2 1 x 2 2 2 3 1 1 g ( x ) x 2 2 3 4 4 x 2 2 g ( x ) 0 1 4 x 2 3 2 _ has _ no _ solutions x 2 _ is _ a _ discontinuity g(x) x 22 1 1 1 x 2 2 1 1 x 2 2 2 2 1 0 _ has _ no _ solutions 1 2 x 2 g(x) x 2 _ is _ a _ discontinuity Practice problem: Find the interval where the function is concave up or concave down. g(x) x 2 The Rule: 1.Find all values of x for which f”(x)=0 or f” is discontinuous and identify the open intervals determined by these points. 2.Select a test point c in each interval found in step 1 and determine the sign of f”(c) in that interval. A. If f”(c) >0 in that interval, f is concave upward on that interval. B. If f”(c) <0 in that interval, f is concave downward on that interval. This is the correct first step. g( x ) x 2 g( x ) x 2 1 2 1 1 1 1 2 x 2 1 x 2 2 2 2 3 1 1 g ( x ) x 2 2 3 4 4 x 2 2 g ( x ) 0 1 4 x 2 3 2 _ has _ no _ solutions x 2 _ is _ a _ discontinuity 1.Find all values of x for which f”(x)=0 or f” is discontinuous and identify the open intervals determined by these points. ,2 is not in the domain of the function (2, ) is the only interval to consider Test point x = 3, g”(3)=-1 The function is concave down on (2, ) Rule 2B y 8 7 6 5 4 3 2 1 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 x 10 This is a typical square root function. It is concave down from (2, ) Inflection Points A point on the graph of a differentiable function f at which the concavity changes is called an inflection point. Finding Inflection Points 1. Compute f”(x) 2. Determine the points in the domain of f for which f”(x) = 0 or f”(x) does not exist. 3. Determine the sign of f”(x) to the left and right of each point x = c found in step 2. If there is a change in the sign of f”(x) as we move across the point x = c, then (c,f(c)) is an inflection point of f. Determine where the function is concave up and concave down and any inflection points for 1 f(x) 2 x 1 Step 1: Find f’ and then f” of x. 1 1 2 x 1 x2 1 f(x) 1(x2 1) 2 2x f(x) simpli fy f(x) 2x(x2 1) 2 f(x) 2x[2(x2 1) 3 2x] (x2 1) 2 2 simpli fy f(x) 2(x2 1) 3 [4x2 x2 1] 2 3x2 1 f(x) 2(x 1) [3x 1] (x2 1)3 2 3 2 2 3x2 1 f(x) (x2 1)3 Step 2: Set f”(x) = 0 and also look for values where x is undefined. 2 3x2 1 0 (x2 1)3 0 2 3x2 1 0 6x2 2 2 6x2 2 x2 6 1 x2 3 1 x 3 The numerator provides us with zeroes. x 1 3 2 3x2 1 f(x) (x2 1)3 The denominator provides us with undefined places. 2 3x2 1 f(x) (x2 1)3 (x2 1)3 0 3 (x2 1)3 3 0 x2 1 0 x2 1 no _ real _ solutions So our only candidates for inflection points are: x 1 3 Step 3: Draw a number line and place the x values on it, then choose test points to the right and left of the values. Result + 1 3 1 3 1 is approximately -.57 3 3 Intervals 1 , 3 1 1 , 3 3 1 , 3 + Test Points -1 0 1 2 3x2 1 f(x) Substituting into (x2 1)3 2(3(1)2 1) 2(3 1) 2(2) 4 1 ((1)2 1)3 23 8 2 1 13 2(3(0)2 1) 2(0 1) 2 2 ((0)2 1)3 0 13 13 2(3(1)2 1) 2(3 1) 2(2) 4 1 ((1)2 1)3 23 8 2 1 13 Interpreting the results On On On 1 , 3 1 1 , 3 3 1 , 3 f is concave up. f is concave down. f is concave up. Let’s look at the graph of f. The blue area is where f is concave up, and the red area is where we are concave down. Since there is a sign change over each x value below, the two points, are inflection points. 1 x Result 3 + 1 3 + 1 3 The Second Derivative Test Because the second derivative tests concavity of a function, it seems like a natural extension to let the second derivative tell us whether we have a relative maximum or relative minimum for a function. Here is how we will do that: 1. Compute f’(x) and f”(x). 2. Find the critical points of f at which f’(x) = 0. 3. Compute f”(c) for each critical point c found in step 2. A. If f”( c) < 0 then f has a relative maximum at c. B. If f”( c) > 0 then f has a relative minimum at c. C. If f”( c) = 0 then f this test fails and you need to go back and use the first derivative test. Why is this true? A. If f”( c) < 0 then f has a relative maximum at c. If f”( c) < 0 then f is concave down which creates a maximum B. If f”( c) > 0 then f has a relative minimum at c. If f”( c) > 0 then f is concave up which creates a minimum Determine the relative extrema for the function: f(x) x4 2x2 4 1. Compute f’(x) and f”(x). f(x) x 4 2x2 4 f(x) 4x3 4x f(x) 12x2 4 2. Find the critical points of f at which f’(x) = 0. f(x) 4x3 4x 0 4x 3 4x 0 4x(x2 1) 0 4x(x 1)(x 1) x 0, x 1, x 1 3. Compute f”(c) for each critical point c found in step 2. c1 = 0, c2 = -1, c3 = 1 Three Critical points. f(x) 12x2 4 2 f(0) 120 4 4 "" 2 f(1) 12 1 4 8 "" 2 f(1) 121 4 8 "" Interpretation of results: A. If f”( c) < 0 then f has a relative maximum at c. B. If f”( c) > 0 then f has a relative minimum at c. C. If f”( c) = 0 then f this test fails and you need to go back and use the first derivative test. f(x) 12x2 4 2 f(0) 120 4 4 "" 2 f(1) 12 1 4 8 "" 2 f(1) 121 4 8 "" Since f”(0) < 0, f has a relative maximum at x = 0. The point is (0, 4) Since f”(-1) > 0, f has a relative minimum at x = -1. The point is (-1, 3) Since f”(1) > 0, f has a relative minimum at x = 1. The point is (1, 3) The graph of f(x) x4 2x2 4 (0,4) rel max (-1,3) rel min (1,3) rel min Go to the HW