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The Derivative in Graphing and Applications By: Tair Akhmejanov.

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Presentation on theme: "The Derivative in Graphing and Applications By: Tair Akhmejanov."— Presentation transcript:

1 The Derivative in Graphing and Applications By: Tair Akhmejanov

2 Section 5.1:Analysis of Functions Increasing Decreasing Concavity Inflection Points

3 To describe functions we need a few definitions A function is increasing on an interval if f(x 1 )f(x 2 ) when x 1 >x 2 A function is constant on an interval if f(x 1 )=f(x 2 ) for all x 1 and x 2 Thus, a function is increasing on an interval where it has a positive slope, decreasing on an interval where it has a negative slopes, and constant on an interval where its graph has zero slope. Let f be a function that is continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If for every value of x in (a,b), then f is increasing on [a,b] If for every value of x in (a,b), then f is decreasing on [a,b] If for every value of x in (a,b), then f is constant on [a,b]

4 If f is differentiable on an open interval I, then f is concave up on I if is increasing on I, and f is concave down on I if is decreasing on I. Let f be differentiable twice on an open interval I. If on I, then f is concave up on I. If on I, then f is concave down on I. If f is continuous on an open interval containing a value x o, and if f changes the direction of its concavity at the point (x o,f(x o )), then we say that f has a n inflection point at x o, and we call the point (x o,f(x o )) on the graph of f an inflection point of f. Concavity

5 State when is increasing, decreasing, concave up, concave down, and all of the inflection points f(x) is decreasing on (-,-3] U [-1, ) f(x) is increasing on [-3,-1] f(x) is concave down on (-2, ) f(x) is concave up on (, -2) f(x) has an inflection point at (-2,-2) Try It

6 State when is increasing, decreasing, concave up, concave down, and all of the inflection points. f(x) is increasing on (-,-8)U(-8, ) f(x) is never decreasing f(x) is concave up on (-, -8) f(x) is concave down on (-8, ) f(x) has no inflection points Try It

7 Section 5.2 Relative Extrema First Derivative Test Second Derivative Test

8 Suppose that f is a function defined on an open interval containing the number x o. If f has a relative extremum at x=x o then either or f is not differentiable at x o. First Derivative Test If on an open interval extending left from x and on an open interval extending right from x o, then f has a relative maximum at x o. If on an open interval extending left from x and on an open interval extending right from x o, then f has a relative minimum at x o. If has the same sign on an open interval extending left from x o and on an open interval extending right from x o, then f does not have a relative extremum at x o. Second Derivative Test If and > 0, then f has a relative minimum at x o. If and < 0, then f has a relative maximum at x o. If and = 0, then the test is inconclusive; that is, f may have a relative maximum, a relative minimum, or neither at x o.

9 Find the relative extremum of Find the relative extremum of By second derivative test f(x) has a relative maximum at (0,0), and relative minimums at (2,-4) and (-2,-4) By second derivative test f(x) has a relative minimum at and a relative maximum at Try It

10 Section 5.4 Rectilinear motion

11 Rectilinear Motion If s(t) is the position function of a particle moving on a coordinate line, then the instantaneous velocity of the particle at time t is defined by: Instantaneous speed at time t: If s(t) is the position function of a particle moving on a coordinate line, then the instantaneous acceleration of the particle at time t is defined by: *Note*: A particle in rectilinear motion is speeding up when its velocity and acceleration have the same sign and slowing down when they have opposite signs.

12 Reminders about particles When the velocity is negative, the particle is moving to the left. When the velocity is negative, the particle is moving to the left. When the velocity is positive, the particle is moving to the right. When the velocity is positive, the particle is moving to the right. When the velocity and acceleration of the particle have the same signs, the particles speed is increasing. When the velocity and acceleration of the particle have the same signs, the particles speed is increasing. When the velocity and acceleration of the particle have opposite signs, the particles speed is decreasing (or slowing down). When the velocity and acceleration of the particle have opposite signs, the particles speed is decreasing (or slowing down). When the velocity is zero and the acceleration is not zero, the particle is momentarily stopped and changing direction. When the velocity is zero and the acceleration is not zero, the particle is momentarily stopped and changing direction.

13 If the position function of a particle is, t>0 find when the particle is changing directions. If the position function of a particle is, t>0 when is the particle speeding up? Velocity is never 0 so particle never changes direction (4, ) Try It

14 Section 5.5 Absolute Maxima and Minima

15 A function f is said to have an absolute maximum on an interval I at x o if f(x o ) is the largest value of f on I; f(x o ) f(x) for all x in the domain of f that are in I. A function f is said to have an absolute maximum on an interval I at x o if f(x o ) is the largest value of f on I; f(x o ) f(x) for all x in the domain of f that are in I. A function f is said to have an absolute minimum on an interval I at x o if f(x o ) is the smallest value of f on I; f(x o ) f(x) for all x in the domain of f that are in I. Extreme-Value Theorem If a function f is continuous on a finite closed interval [a,b], then f has both an absolute maximum and an absolute minimum on [a,b]. If a function f is continuous on a finite closed interval [a,b], then f has both an absolute maximum and an absolute minimum on [a,b]. If f has an absolute extremum on an open interval (a,b), then it must occur at a critical number of f. If f has an absolute extremum on an open interval (a,b), then it must occur at a critical number of f. *Note*: Calculate all critical values of f and calculate values at endpoints. *Note*: Calculate all critical values of f and calculate values at endpoints. Largest value is the absolute maximum and smallest value is the absolute minimum. Largest value is the absolute maximum and smallest value is the absolute minimum. Suppose f is continuous and has exactly one relative extremum on an interval I at x o. Suppose f is continuous and has exactly one relative extremum on an interval I at x o. If f has a relative minimum at x o, then f(x o ) is the absolute minimum of f on I. If f has a relative maximum at x o, then f(x o ) is the absolute maximum of f on I.

16 At what value of x does f(x) have an absolute minimum given that At what point does f(x) have an absolute minimum and at what point does f(x) have an absolute maximum given that on the interval 4x9 There is an absolute minimum at (5.24,10.47) Try It By the second derivative test f(x) has an absolute minimum at

17 Section 5.6 Applied Maximum and Minimum Problems

18 Suggestions for Applied Maximum and Minimum Problems Draw a figure and label the quantities relevant to the problem. Find a formula for the quantity to be maximized or minimized Using the conditions stated in the problem to eliminate variables, express the quantity to be maximized or minimized as a function of one variable. Find the interval of possible values for this variable from the physical restrictions in the problem. If applicable, use the techniques of the preceding section to obtain the maximum or minimum. Note: Profit = Revenue - Cost P(x) = R(x) - C(x) P(x) = R(x) - C(x)

19 Find the point on closest to point (4,0) Use the distance formula. Plug in for y. Let L(x)=D^2 because the minimum value of D^2 will occur at the same value of x as the minimum value of D. Try It

20 Given a rectangular sheet of cardboard 18 inches by 24 inches a box is made by cutting a square with sides x inches from each corner and bending the resulting sides up. Find the value of x that maximizes the volume of the box. Width: 18-2x Length: 24-2x Depth: x Volume=x(18-2x)(24-2x) Volume of the box maximized at x=3.4 Try It Eliminate x=10.6 because it does not work in the context of the problem.

21 Section 5.7 Newtons Method, n=1,2,3…

22 Newtons Method Newtons Method is used to find the point at which the graph of a function intersects the x-axis. Usually you are given a starting value, allowed to use a calculator to pick a starting value, or required to make an educated guess about the root to pick a good starting value.

23 Given a function, x>0 find its root using Newtons Method. Start at x 1 =2.1 Given a function, x>0 find its root using Newtons Method (you may use a calculator to get your starting value for x) The more times this method is repeated by plugging in the new value for x, the more accurate the approximation of the root will be. In this case the root is x=2. X= A question will usually tell you to run the test a certain amount of times. Try It

24 Section 5.8 Mean-Value Theorem Rolles Theorem

25 The Mean Value Theorem (for derivatives) Let f be differentiable on (a,b) and continuous on [a,b]. Then there is at least one number c in (a,b) such that: Another way of putting it: Given a function f there is a point c in an interval (a,b) where the slope of the tangent line equals the slope of the secant line of the endpoints of the interval.

26 Another way of putting it : The Rolles theorem is just an instance of the Mean Value Theorem when the slope is zero, and therefore the tangent line is a horizontal tangent. Let f be differentiable on (a,b) and continuous on [a,b]. If f(a)=f(b)=0, then there is at least one number c in (a,b) such that f(c)=0. Rolles Theorem

27 on the interval [-4,4] Find the values of c that satisfy the Mean Value Theorem Try It

28 Find values of c that satisfy the Mean Value Theorem for on the interval [-1,2] Find the values of c that satisfy Rolles Theorem for on the interval [0,1] No Solution C=1/2

29 1) Find the values of c that satisfy Rolles Theorem for on the interval [-1,1] 2) Find the absolute maximum and minimum values of on the interval [-3,3] 3) A rectangle is inscribed in a semicircle, that has a radius of 4, with one side on the semicircles diameter. What is the largest area this rectangle can have? A) QUIZQUIZ B)C) D) A) 16 B) C) -4 D) 9 A)min.=-3 B) min.= C) min=-24 D) min.=-.385 max=3 max= max=24 max=.385

30 4) A rectangular field, bounded on one side by a building, is to be fenced in on the other three sides. If 3,000 feet of fence is to be used, find the dimensions of the largest field that can be fenced in. 5) Find the values of c that satisfy the Mean Value Theorem for on the interval [0,4] 6) Find the point at the relative maximum and the point at the relative minimum of A) 750 ft. by 1,000 ft. B)500 ft. by 1,500 ft. C)350 ft. by 2,300 ft. D)750 ft. by 1,500 ft. A) B) C) D) A)max.:(4,0) B)no relative maximum C)max:(3,8) D)max:(5,24) min:(3,-27) min:(3,-27) no relative minimum min:(-1,-14)

31 7) A poster is to contain 100 square inches of picture surrounded by a 4-inch margin at the top and bottom and a 2-inch margin on each side. Find the overall dimensions that will minimize the total area of the poster. 8) Given the position function, find when the particle is speeding up. 9) If the position function of a particle is 0t>7 A)t=,3 B)t= /2 C)t=, /2 D)4

32 10) Given a function state when it is concave up, concave down, and the points of inflection. A)Cocave up: (-3,0)U(2, ) Concave down: (-,-3)U(0,2) Inflection point(s): (-3,189); (0,0); (2, -16) D) Concave up: (-,0)U(2, ) Concave down: (0,2) Inflection point(s): (2,-16) B) Concave up: (-, ) Concave down: never Inflection point(s): none C) Concave up: (-,0)U(2, ) Concave down: (0,2) Inflection point(s): (2,0)

33 Credits This Presentation: Tair Akhmejanov Fact Checker: Alex Bishop Text Book: Anton, Bivens, Davis Princeton Review: David S. Kahn Barrons: Shirley O. Hockett, David Bock PowerPoint: Microsoft Corp.

34 Bibliography of Pictures


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