Calculus Date: 12/18/13 Obj : SWBAT apply first derivative test http ://youtu.be/PBKnttVMbV4 first derivative test inc. dec. Today – I Cover the second derivative test (easier than 1 st ) https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/

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Calculus Date: 12/18/13 Obj : SWBAT apply first derivative test

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Calculus Date:12/18/13 Obj: SWBAT apply first derivative test http://youtu.be/PBKnttVMbV4 first derivative test inc.dec. Today –ICover the second derivative test (easier than 1st) https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/ n class: Start WS3-3A; complete for homework Tomorrow Friday: With Christian complete any remaining worksheets, make sure You understand material from the test.,etc. • Announcements: • Break Packet online on Friday • Merry Christmas if I don’t see you "Do not judge me by my successes, judge me by how many times I fell down and got back up again.“ Nelson Mandela

Second Derivative Test The first derivative test tells whether the function, f(x), is increasing, decreasing or constant. The second derivative test tells whether the derivative, f’(x),is increasing, decreasing or constant, or whether a function is concave up or concave down and whether a point is a local max or min.

Concave upward y y x x • Look at these two graphs. Each is concave upward, but one is decreasing and the other is increasing. We need to be able to determine concavity from the function and not just from the graph.

Concave downward y y x x • Look at these two graphs. Each is concave downward, but one is decreasing and the other is increasing. We need to be able to determine concavity from the function and not just from the graph.

Concavity Let f be a differentiable function on (a, b). 1.f is concave upward on (a, b) if f' is increasing on aa(a, b). That is f ''(x)0 for each value of x in (a, b). 2.f is concave downward on (a, b) if f' is decreasing on (a, b). That is f ''(x)0 for each value of x in (a, b). concave upward concave downward

Inflection Point A point on the graph of f at which fis continuousandconcavity changes is called an inflection point. To search for inflection points, find any point, c in the domain where f ''(x)0 or f ''(x)is undefined. Some call these points “hypercritical points”. If f ''changes sign from the left to the right of c, then (c,f (c))is an inflection point of f.

The Second Derivative Test Determine the sign of the second derivative of f to the left and right of the hypercritical point to determine if the function is concave up or concave down or neither. left right conclusion f (c) is an inflection point up down down up f (c) is an inflection point No change Concavity remains the same

Relative min. f (4) = -31 Relative max. f (0) = 1 The First and Second Derivative Test Graph and find all the relative extrema of = 0 x = 2 - 2 + + 0 - 0 + 0 4 f(x) is concave down for all x in (-∞, 2) because f” < 0. There is an Inflection point at (2, -15) f(x) = 0 and f(x) becomes concave upward

The First and Second Derivative Tests f(x) is concave down for all x in (-∞, 2) because f” < 0. There is an inflection point at (2, -15) and f(x) =0 and where f(x) becomes concave upward What else can we say about the rate of change? There is a local maximum at (0,1)because >0 for all x in (-∞, 0) and < 0 for all x in (0,2). There is a local minimum at (4,-31)because <0 for all x in (0,4)and > 0 for all x in (4, ∞).

Also the Second Derivative Test can be used to find local extrema Let c be a critical point of f(x). If f”(c) exists, then • If f ''(c) > 0 then f(c) is a local minimum. • If f ''(c) < 0 then f(c) is a local maximum. • If f ''(c) = 0 inconclusive, use the first derivative test to classify the extrema.

Second Derivative Test Let c be a critical point of f(x). If f”(c) exists, then • If f ''(c) > 0 then f(c) is a local minimum. • If f ''(c) < 0 then f(c) is a local maximum. • If f ''(c) = 0 inconclusive, use the first derivative test to classify the extrema. Using the function below, find any relative extrema f(0)= -12 f(4) = 12

Graph • Take the derivative f’(x) • Find the critical points f’(x) = 0; f’(x) = DNE • Make sign chart • Label critical points – put 0 or DNE on graph • Evaluate the derivative at points on either side of extrema to determine the sign. Use Yvars for faster calculations. Mark the chart as + or – in these areas • Evaluate f(x) at the critical points to determine actual values of extrema. • Add arrows to show increasing or decreasing regions in f(x) • Write out all extrema and use the value of the derivative and the appropriate interval to justify your answer.

The Point of Diminishing Returns If the function represents the total sales of a particular object, t months after being introduced, find the point of diminishing returns. S concave up on S concave down on The point of diminishing returns is at 20 months (the rate at which units are sold starts to drop).

A similar Observation Applies at a Local Max. Generic Example The First Derivative Test

Another Example Find all the relative extrema of Stationary points: Singular points:

Stationary points: Singular points: Evaluate the derivative at points on either side of extrema to determine the sign. Use Yvars for faster calculations ND – derivative not defined Relative max. Relative min. + ND + 0 - ND - 0 + ND + -1 0 1

Local min. Local max. Graph of + ND + 0 - ND - 0 + ND + -1 0 1

There are roots at and . Possibleextremaat . Set Example: Graph We can use a chart to organize our thoughts. First derivative test: y y=4 y=0 positive negative positive

There are roots at and . Possible extreme at . Set maximum at minimum at Example: Graph First derivative test: y y=4 y=0

There is a local maximum at (0,4) because for all x in and for all x in (0,2) . There is a local minimum at (2,0) because for all x in (0,2) and for all x in . Example: Graph NOTE: On the AP Exam, it is not sufficient to simply draw the chart and write the answer. You must give a written explanation! First derivative test:

There is a local maximum at (0,4) because for all x in and for all x in (0,2) . There is a local minimum at (2,0) because for all x in (0,2) and for all x in . Example: Graph First derivative test:

Applications of the Derivative • Maxima and Minima • Applications of Maxima and Minima • The Second Derivative - Analyzing Graphs

Absolute Extrema Let f be a functiondefined on a domain D Absolute Maximum Absolute Minimum

Absolute Extrema A function f has an absolute (global) maximum atx = c if f (x) f (c)for allx in the domain D of f. The number f (c) is called the absolute maximumvalue of f in D Absolute Maximum

Absolute Extrema A function f has an absolute (global) minimum atx = c if f (c) f (x)for allx in the domain D of f. The number f (c) is called the absolute minimumvalue of f in D Absolute Minimum

Relative Extrema A function f has a relative (local) maximum at xc if there exists an open interval (r, s) containing c suchthat f (x) f (c) for all r x s. Relative Maxima

Relative Extrema A function f has a relative (local) minimum at xc if there exists an open interval (r, s) containing c suchthat f (c) f (x) for all r x s. Relative Minima

Generic Example The corresponding values of x are called Critical Points of f

Critical Points of f A critical number of a function f is a number cin the domain off such that (stationary point) (singular point)

Candidates for Relative Extrema • Stationary points: any x such that xis in the domain of f and f'(x) 0. • Singular points: any x such that xis in the domain of f and f'(x) undefined • Remark:notice that not every critical number correspond to a local maximum or local minimum. We use “local extrema” to refer to either a max or a min.

Fermat’s Theorem If a function f has a local maximum or minimum at c, then c is a critical number of f Notice that the theorem does not say that at every critical number the function has a local maximum or local minimum

Generic Example Two critical points of f that do not correspond to local extrema

Example Find all the critical numbers of Stationary points: Singular points:

Extreme Value Theorem If a function f is continuous on a closed interval [a,b], then f attains an absolute maximum and absolute minimum on [a, b]. Each extremum occurs at a critical number or at an endpoint. a b a b a b Attains max. and min. Attains min. but no max. No min. and no max. Open Interval Not continuous