4.1 Extreme Values of Functions Absolute extreme values are either maximum or minimum points on a curve. They are sometimes called global extremes. They are also sometimes called absolute extrema. (Extrema is the plural of the Latin extremum.)

4.1 Extreme Values of Functions Definition Absolute Extreme Values Let f be a function with domain D. Then f (c) is the absolute minimum value on D if and only if f(x) < f (c) for all x in D. absolute maximum value on D if and only if f(x) > f (c) for all x in D.

4.1 Extreme Values of Functions Extreme values can be in the interior or the end points of a function. No Absolute Maximum Absolute Minimum

4.1 Extreme Values of Functions Absolute Maximum Absolute Minimum

4.1 Extreme Values of Functions Absolute Maximum No Minimum

4.1 Extreme Values of Functions No Maximum No Minimum

4.1 Extreme Values of Functions Extreme Value Theorem: If f is continuous over a closed interval, [a,b] then f has a maximum and minimum value over that interval. Maximum & minimum at interior points Maximum & minimum at endpoints Maximum at interior point, minimum at endpoint

4.1 Extreme Values of Functions Local Extreme Values: A local maximum is the maximum value within some open interval. A local minimum is the minimum value within some open interval.

4.1 Extreme Values of Functions Absolute maximum (also local maximum) Local maximum Local minimum Local minimum Local extremes are also called relative extremes. Absolute minimum (also local minimum)

4.1 Extreme Values of Functions Absolute maximum (also local maximum) Local maximum Local minimum Notice that local extremes in the interior of the function occur where is zero or is undefined.

4.1 Extreme Values of Functions Local Extreme Values: If a function f has a local maximum value or a local minimum value at an interior point c of its domain, and if exists at c, then

4.1 Extreme Values of Functions Critical Point: A point in the domain of a function f at which or does not exist is a critical point of f . Note: Maximum and minimum points in the interior of a function always occur at critical points, but critical points are not always maximum or minimum values.

4.1 Extreme Values of Functions EXAMPLE 3 FINDING ABSOLUTE EXTREMA Find the absolute maximum and minimum values of on the interval . There are no values of x that will make the first derivative equal to zero. The first derivative is undefined at x=0, so (0,0) is a critical point. Because the function is defined over a closed interval, we also must check the endpoints.

4.1 Extreme Values of Functions To determine if this critical point is actually a maximum or minimum, we try points on either side, without passing other critical points. At: Since 0<1, this must be at least a local minimum, and possibly a global minimum. At: At:

4.1 Extreme Values of Functions Absolute minimum: maximum: At: At:

4.1 Extreme Values of Functions y = x2/3

4.1 Extreme Values of Functions Finding Maximums and Minimums Analytically: 1 Find the derivative of the function, and determine where the derivative is zero or undefined. These are the critical points. 2 Find the value of the function at each critical point. 3 Find values or slopes for points between the critical points to determine if the critical points are maximums or minimums. 4 For closed intervals, check the end points as well.

4.1 Extreme Values of Functions Find the absolute maximum and minimum of the function Find the critical numbers

4.1 Extreme Values of Functions Find the absolute maximum and minimum of the function Check endpoints and critical numbers The absolute maximum is 2 when x = -2 The absolute minimum is -13 when x = -1

4.1 Extreme Values of Functions Find the absolute maximum and minimum of the function Find the critical numbers

4.1 Extreme Values of Functions Find the absolute maximum and minimum of the function Check endpoints and critical numbers The absolute maximum is 3 when x = 0, 3 The absolute minimum is 2 when x = 1

4.1 Extreme Values of Functions Find the absolute maximum and minimum of the function Find the critical numbers

4.1 Extreme Values of Functions Find the absolute maximum and minimum of the function The absolute maximum is 1/4 when x = /6, 5/6 The absolute minimum is –2 when x =3/2

4.1 Extreme Values of Functions Critical points are not always extremes! (not an extreme)

4.1 Extreme Values of Functions (not an extreme)

4.2 Mean Value Theorem Mean Value Theorem for Derivatives If f (x) is a differentiable function over [a,b], then at some point between a and b: Mean Value Theorem for Derivatives

4.2 Mean Value Theorem Mean Value Theorem for Derivatives If f (x) is a differentiable function over [a,b], then at some point between a and b: Mean Value Theorem for Derivatives Differentiable implies that the function is also continuous.

4.2 Mean Value Theorem Mean Value Theorem for Derivatives If f (x) is a differentiable function over [a,b], then at some point between a and b: Mean Value Theorem for Derivatives Differentiable implies that the function is also continuous. The Mean Value Theorem only applies over a closed interval.

4.2 Mean Value Theorem Mean Value Theorem for Derivatives If f (x) is a differentiable function over [a,b], then at some point between a and b: Mean Value Theorem for Derivatives The Mean Value Theorem says that at some point in the closed interval, the actual slope equals the average slope.

4.2 Mean Value Theorem Tangent parallel to chord. Slope of tangent: Slope of chord:

4.2 Mean Value Theorem Rolle’s Theorem If f (x) is a differentiable function over [a,b], and if f(a) = f(b) = 0, then there is at least one point c between a and b such that f’(c)=0: (a,0) (b,0)

4.2 Mean Value Theorem Show the function satisfies the hypothesis of the Mean Value Theorem The function is continuous on [0,/3] and differentiable on (0,/3). Since f(0) = 1 and f(/3) = 1/2, the Mean Value Theorem guarantees a point c in the interval (0,/3) for which c = .498

4.2 Mean Value Theorem at x = .498, the slope of the tangent line is (0,1) at x = .498, the slope of the tangent line is equal to the slope of the chord. (/3,1/2)

4.2 Mean Value Theorem Definitions Increasing Functions, Decreasing Functions Let f be a function defined on an interval I and let x1 and x2 be any two points in I. f increases on I if x1 < x2  f(x1) < f(x2). f decreases on I if x1 > x2  f(x1) > f(x2).

4.2 Mean Value Theorem Corollary Increasing Functions, Decreasing Functions Let f be continuous on [a,b] and differentiable on (a,b). If f’ > 0 at each point of (a,b), then f increases on [a,b]. If f’ < 0 at each point of (a,b), then f decreases on [a,b]. A couple of somewhat obvious definitions: A function is increasing over an interval if the derivative is always positive. A function is decreasing over an interval if the derivative is always negative.

4.2 Mean Value Theorem Find where the function is increasing and decreasing and find the local extrema. + - 2 4 f’(x) x = 2, local maximum x = 4, local minimum

4.2 Mean Value Theorem (2,20) local max (4,16) local min

4.2 Mean Value Theorem These two functions have the same slope at any value of x. Functions with the same derivative differ by a constant.

4.2 Mean Value Theorem Find the function whose derivative is and whose graph passes through so:

4.2 Mean Value Theorem Find the function f(x) whose derivative is sin(x) and whose graph passes through (0,2). so: Notice that we had to have initial values to determine the value of C.

4.2 Mean Value Theorem The process of finding the original function from the derivative is so important that it has a name: Antiderivative A function is an antiderivative of a function if for all x in the domain of f. The process of finding an antiderivative is antidifferentiation. You will hear much more about antiderivatives in the future. This section is just an introduction.

4.2 Mean Value Theorem Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward. (We let down be positive.) Since acceleration is the derivative of velocity, velocity must be the antiderivative of acceleration.

4.2 Mean Value Theorem The power rule in reverse: Increase the exponent by one and multiply by the reciprocal of the new exponent. Since velocity is the derivative of position, position must be the antiderivative of velocity.

4.2 Mean Value Theorem The initial position is zero at time zero.

4.3 Connecting f’ and f’’ with the Graph of f In the past, one of the important uses of derivatives was as an aid in curve sketching. We usually use a calculator of computer to draw complicated graphs, it is still important to understand the relationships between derivatives and graphs.

4.3 Connecting f’ and f’’ with the Graph of f First Derivative Test for Local Extrema at a critical point c If f ‘ changes sign from positive to negative at c, then f has a local maximum at c. local max f’>0 f’<0 2. If f ‘ changes sign from negative to positive at c, then f has a local minimum at c. local min f’<0 f’>0 If f ‘ changes does not change sign at c, then f has no local extrema. no extreme f’>0

4.3 Connecting f’ and f’’ with the Graph of f First derivative: is positive Curve is rising. is negative Curve is falling. is zero Possible local maximum or minimum.

4.3 Connecting f’ and f’’ with the Graph of f concave up Definition Concavity The graph of a differentiable function y = f(x) is concave up on an open interval I if y’ is increasing on I. (y’’>0) concave down on an open interval I if y’ is decreasing on I. (y’’<0) concave down

4.3 Connecting f’ and f’’ with the Graph of f Second Derivative Test for Local Extrema at a critical point c If f’(c) = 0 and f’’(c) < 0, then f has a local maximum at x = c. If f’(c) = 0 and f’’(c) > 0, then f has a local minimum at x = c. + +

4.3 Connecting f’ and f’’ with the Graph of f Second derivative: is positive Curve is concave up. is negative Curve is concave down. is zero Possible inflection point (where concavity changes).

4.3 Connecting f’ and f’’ with the Graph of f Definition Point of Inflection A point where the graph of a function has a tangent line and where the concavity changes is called a point of inflection. inflection point

4.3 Connecting f’ and f’’ with the Graph of f Sketch the graph Possible extreme at . zeros at x = -1, x = 2 First derivative test: Set negative positive positive

4.3 Connecting f’ and f’’ with the Graph of f Possible extreme at . Set First derivative test: maximum at minimum at

4.3 Connecting f’ and f’’ with the Graph of f Possible extreme at . Set Or you could use the second derivative test: negative concave down local maximum positive concave up local minimum maximum at minimum at

4.3 Connecting f’ and f’’ with the Graph of f We then look for inflection points by setting the second derivative equal to zero. Possible inflection point at . negative positive inflection point at

4.3 Connecting f’ and f’’ with the Graph of f Make a summary table: rising, concave down local max falling, inflection point local min rising, concave up

4.4 Modeling and Optimization A Classic Problem You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?

4.4 Modeling and Optimization

4.4 Modeling and Optimization To find the maximum (or minimum) value of a function: Understand the Problem. Develop a Mathematical Model. Graph the Function. Identify Critical Points and Endpoints. Solve the Mathematical Model. Interpret the Solution.

4.4 Modeling and Optimization What dimensions for a one liter cylindrical can will use the least amount of material? Motor Oil We can minimize the material by minimizing the area. We need another equation that relates r and h: area of ends lateral area

4.4 Modeling and Optimization area of ends lateral area

4.4 Modeling and Optimization Find the radius and height of the right-circular cylinder of largest volume that can be inscribed in a right-circular cone with radius 6 in. and height 10 in. 10 in r h 6 in

4.4 Modeling and Optimization The formula for the volume of the cylinder is h r 10 in 6 in To eliminate one variable, we need to find a relationship between r and h. 6 h 10-h r 10

4.4 Modeling and Optimization h r 10 in 6 in

4.4 Modeling and Optimization h r 10 in 6 in Check critical points and endpoints. r = 0, V = 0 r = 4 V = 160/3 r = 6 V = 0 The cylinder will have a maximum volume when r = 4 in. and h = 10/3 in.

4.4 Modeling and Optimization Determine the point on the curve y = x2 that is closest to the point (18, 0).                 Substitute for x

4.4 Modeling and Optimization Determine the point on the curve y = x2 that is closest to the point (18, 0).                

4.4 Modeling and Optimization Determine the point on the curve y = x2 that is closest to the point (18, 0).                 - 0 + 2

4.4 Modeling and Optimization Notes: If the function that you want to optimize has more than one variable, use substitution to rewrite the function. If you are not sure that the extreme you’ve found is a maximum or a minimum, you have to check. If the end points could be the maximum or minimum, you have to check.

4.5 Linearization and Newton’s Method For any function f (x), the tangent is a close approximation of the function for some small distance from the tangent point. We call the equation of the tangent the linearization of the function.

4.5 Linearization and Newton’s Method Start with the point/slope equation: linearization of f at a is the standard linear approximation of f at a. The linearization is the equation of the tangent line, and you can use the old formulas if you like.

4.5 Linearization and Newton’s Method Find the linearization of f(x) = x4 + 2x at x = 2 f’(x) = 4x3 + 2 L (x) = f(3) + f’(3)(x - 3) L (x) = 87 + 110(x - 3) L (x) = 110x - 243

4.5 Linearization and Newton’s Method Important linearizations for x near zero: This formula also leads to non-linear approximations:

4.5 Linearization and Newton’s Method Estimate using local linearization.

4.5 Linearization and Newton’s Method Estimate sin 31º using local linearization. Need to be in radians

4.5 Linearization and Newton’s Method Differentials: When we first started to talk about derivatives, we said that becomes when the change in x and change in y become very small. dy can be considered a very small change in y. dx can be considered a very small change in x.

4.5 Linearization and Newton’s Method Let y = f(x) be a differentiable function. The differential dx is an independent variable. The differential dy is: dy = f ’(x)dx

4.5 Linearization and Newton’s Method Example: Consider a circle of radius 10. If the radius increases by 0.1, approximately how much will the area change? very small change in r very small change in A (approximate change in area)

4.5 Linearization and Newton’s Method Compare to actual change: New area: Old area: Absolute error percent error

4.5 Linearization and Newton’s Method True Estimated Absolute Change Relative Change Percent Change

4.5 Linearization and Newton’s Method x y y = f(x) Root sought x1 First (x1,f(x1)) (x2,f(x2)) (x3,f(x3)) x3 Third x2 Second

4.5 Linearization and Newton’s Method This is Newton’s Method of finding roots. It is an example of an algorithm (a specific set of computational steps.) This is a recursive algorithm because a set of steps are repeated with the previous answer put in the next repetition. Each repetition is called an iteration.

4.5 Linearization and Newton’s Method Finding a root for: We will use Newton’s Method to find the root between 2 and 3.

4.5 Linearization and Newton’s Method Guess x1 = 2

4.5 Linearization and Newton’s Method Guess x2 = 2.5

4.5 Linearization and Newton’s Method Find where crosses .

4.5 Linearization and Newton’s Method

4.5 Linearization and Newton’s Method There are some limitations to Newton’s Method: Looking for this root. Bad guess. Wrong root found Failure to converge

4.6 Related Rates First, a review problem: Consider a sphere of radius 10 cm. If the radius changes 0.1 cm (a very small amount) how much does the volume change? The volume would change by approximately 40 cm3 .

4.6 Related Rates Now, suppose that the radius is changing at an instantaneous rate of 0.1 cm/sec. The sphere is growing at a rate of 40 cm3/sec . Note: This is an exact answer, not an approximation like we got with the differential problems.

4.6 Related Rates Water is draining from a cylindrical tank at 3 liters/second. How fast is the surface dropping? (We need a formula to relate V and h. ) Find (r is a constant.)

4.6 Related Rates Steps for Related Rates Problems: 1. Draw a picture (sketch). 2. Write down known information. 3. Write down what you are looking for. 4. Write an equation to relate the variables. 5. Differentiate both sides with respect to t. 6. Evaluate.

4.6 Related Rates Hot Air Balloon Problem: Given: How fast is the balloon rising? Find

4.6 Related Rates Hot Air Balloon Problem: Given: How fast is the balloon rising? Find

4.6 Related Rates Truck Problem: Truck A travels east at 40 mi/hr. Truck B travels north at 30 mi/hr. How fast is the distance between the trucks changing 6 minutes later? B A

4.6 Related Rates Truck Problem: Truck A travels east at 40 mi/hr. Truck B travels north at 30 mi/hr. How fast is the distance between the trucks changing 6 minutes later? B A

4.6 Related Rates Truck Problem: Truck A travels east at 40 mi/hr. Truck B travels north at 30 mi/hr. How fast is the distance between the trucks changing 6 minutes later?