Bob Brown Math 251 Calculus 1 Chapter 4, Section 4 1 CCBC Dundalk

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1 Bob Brown Math 251 Calculus 1 Chapter 4, Section 4 1 A Function and its Second Derivative Recall page 4 of Handout 3.1 where we encountered the third degree polynomial f(x) = x 3 5x 2 4x Its derivative function (which we computed using the limit definition your complaints are still ringing in my ears!) is f (x) = 3x 2 10x 4. Exercise 1: Compute the derivative of f (x) = 3x 2 10x 4 in order to produce the second derivative function of f. Then complete the sentences pertaining to the sign of and the concavity of f. d = ( 3x 2 10x 4) f (x) dx = > 0 and f is concave up on < 0 and f is concave down on Test for Concavity Theorem: Let f be a function whose second derivative exists on an open interval I. 1. f is concave up f (x) is f (x ) is 2. f is concave down f (x) is f (x ) is Def.: The point (c, f(c)) is an inflection point of f if f has a tangent line at x = c and if f changes concavity (from up to down or from down to up) at x = c.

2 Bob Brown Math 251 Calculus 1 Chapter 4, Section 4 2 Exercise 2: Determine the inflection point(s) of f(x) = x 5 2x 4. First, determine the second derivative: f (x) = f (x ) = Next, find the x-values of the candidate inflection point(s) by solving f (x ) = 0. 20x 3 24x 2 = 0 x =, x = 6 Then, use a sign chart to determine whether or not the sign of f (x ) changes at 0 or at. 5 test x: 0 6/5 : sign of f (x ) : From the f (x ) information, we conclude that there is an inflection point at x = Note that the inflection point is , f =, * * The sign chart also grants us the following information. f is concave down on and concave up on Important Note: Not every point (c, f(c)) at which f (c ) = 0 or at which f (c ) is undefined is an inflection point. We just saw this in Exercise 2 at x = 0. As another example, you should 4 verify on your own that the second derivative of f ( x) = x equals 0 at x = 0, but there is not an inflection point at x = 0, since the concavity of f does not change at x = 0.

3 Bob Brown Math 251 Calculus 1 Chapter 4, Section 4 3 Exercise 3a: From the given graph of f, determine the intervals on which f, f, and f are positive or negative. Note that the domain of f is < x <. f(x) > 0 on f(x) < 0 on f (x) > 0 on f (x) < 0 on > 0 on < 0 Exercise 3b: From the given graph of g, determine the intervals on which g, g, and g are positive or negative. Note that the domain of g is 1 < x <. g(x) > 0 on g(x) < 0 on g (x) > 0 on g (x) < 0 g (x) > 0 g (x) < 0 on

4 Bob Brown Math 251 Calculus 1 Chapter 4, Section 4 4 Exercise 3c: From the given graph of h, determine the intervals on which h, h, and h are positive or negative. Note that the domain of h is < x <. Note also that the limitation of the graphing calculator incorrectly portrays h as being constant on a certain interval of its domain. h(x) > 0 on h(x) < 0 on h (x) > 0 on h (x) < 0 h (x) > 0 on h (x) < 0 on The Second Derivative Test for Relative Maxima and Minima Theorem: Let f be a function such that f (c) = 0 and such that f (x ) exists on an open interval containing c. 1. If f (c ) > 0, then f has a at x = c. c 2. If f (c ) < 0, then f has a at x = c. c 3. If f (c ) = 0, then the test fails. That is, f may have a relative maximum, minimum, or neither. In such cases, you should fall back on the First Derivative Test.

5 Bob Brown Math 251 Calculus 1 Chapter 4, Section 4 5 Exercise 4: Recall Exercise 1 on page 2 of Handout 4.3, where we determined that the critical numbers of g(x) = 2x 3 + 3x 2 36x + 1 are x = -3 and x = 2. Use the Second Derivative Test to classify them (as relative maximums, minimums, or neither.) g (x) = g (x ) = Thus, g ( 3 ) = there is a relative at x = -3. and g (2 ) = there is a relative at x = 2. An Example Where the Second Derivative Test Fails Exercise 5: Determine the critical point of f(x) = x 4, and use the Second Derivative Test to classify it, if possible. If the Second Derivative Test fails, fall back on the First Derivative Test. f(x) = x 4 f (x) =. Solving f (x) = Now, f (x) = 3 4x = 0 for x, we get x =. Hence, the critical number is x = 0, and the critical point is (0, f(0)) =. 3 4x ) f (x =. Evaluating the critical number in f (x ), we get f (0 ) = =. Thus, the Second Derivative Test fails tells us nothing. So, we fall back on the First Derivative Test (using a sign chart.)

6 Bob Brown Math 251 Calculus 1 Chapter 4, Section 4 6 Exercise 6: Determine (and classify) the critical point of and determine the inflection point of f(x) = xe -x.