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

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1 Bob Brown Math 251 Calculus 1 Chapter 4, Section 4 Completed 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 f (x) and the concavity of f. d f (x) 3x 2 10x 4 dx f (x) > 0 and f is concave up on f (x) < 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 Completed 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 f (x) : 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 Completed 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 f (x) > 0 on f (x) < 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 Completed 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 Completed 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 (x) f. 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 Completed 6 Exercise 6: Determine (and classify) the critical point of and determine the inflection point of f(x) xe -x.

4.3 Finding Local Extreme Values: First and Second Derivatives

Arkansas Tech University MATH 2914: Calculus I Dr. Marcel B. Finan 4.3 Finding Local Extreme Values: First and Second Derivatives Recall that a function f(x) is said to be increasing (respectively decreasing)