5.1 Maxima and Minima Relative Extrema relative maxima : x values of p, r, b relative minima : x values of a, q, s Absolute Extrema absolute maximum : at x = b absolute minimum : at x = s f has a relative maximum at x = r if there is some interval (r – h, r + h) for which f (r) ≥ f (x) for all x in (r – h, r + h) for which f (x) is defined. f has a relative minimum at x = r if there is some interval (r – h, r + h) for which f (r) ≤ f (x) for all x in (r – h, r + h) for which f (x) is defined. f has an absolute maximum at r if f (r) ≥ f (x) for every x in the domain of f. f has an absolute minimum at r if f (r) ≤ f (x) for every x in the domain of f.

Maxima and Minima Types of extrema Stationary Points where f’(x) = 0 [rate of change of f is 0 there] Singular points exist when f (x) does not exist for some x in the domain (points where there is no horizontal tangent) Endpoints are also considered types of extrema ** Note : Stationary points & singular points are also considered “critical points” At the points labeled “Stationary,” tangent lines to the graph are horizontal (with slope = 0). So, f’ (which gives the slope) is 0.

Finding Extrema Stationary Points: where x is in the domain and f (x) = 0. To locate, set f (x) = 0 and solve for x. Singular Points: where x is in the domain and f (x) is not defined. To locate, find values of x where f (x) is not defined, but f (x) is defined. Endpoints: The x-coordinates of endpoints are endpoints of the domain, if any. (Note that closed intervals have endpoints; open intervals do not). Example1: (Finding stationary points) Let f (x) = x3 – 12x f ’(x) = 3x2 – 12 3x2 – 12 = 0 x = ±2. Stationary points are:(2, -16) & (-2, 16) Example 2: (Finding singular points) Let f (x) = 3(x – 1)1/3 f (x) = (x – 1)−2/3 = 1/(x – 1)2/3 f ‘ (1) is not defined, but f (1) is defined Singular point occurs at x = 1 (1, 0) Example 3: (Endpoints) Let f (x) = 1/x ( -∞ , 0)  [1, ∞ ). The only endpoint in the domain of f occurs when x = 1 and has coordinates (1, 1). The natural domain of 1/x has no endpoints

Another Example Find the relative and absolute maxima and minima of f (x) = x2 – 2x on the interval [0, 4]. Use the derivative to locate the critical points (Stationary and Singular) f (x) = 2x – 2. Check for Stationary Points: Let f ‘(x) = 0 => 2x – 2 = 0 => x = 1 So, x = 1 is Stationary relative extrema point : (1, -1) Check for Singular Points: f ‘ (x) is 2x -2 which is defined for all x on the domain of [0, 4]. Thus, there are no singular points. (Remember singular points must be defined for f(x) but not for f ‘(x) ) Check for End points: x = 0 and x = 4. The points are :(0, 0) and (4, 8) Now, try to draw the graph using the 3 critical points you found and note that you have a relative max, absolute min and absolute max

First Derivative Test for Relative Extrema Suppose that c is a critical point of the continuous function f, and that its derivative is defined for x close to, and on both sides of, x = c. Then, determine the sign of the derivative to the left and right of x = c. 1. If f (x) is positive to the left of x = c and negative to the right, then f has a relative maximum at x = c. 2. If f (x) is negative to the left of x = c and positive to the right, then f has a relative minimum at x = c. 3. If f (x) has the same sign on both sides of x = c, then f has neither a relative maximum nor a relative minimum at x = c. Note the graph has singular point and relative maximum at x = 1

Example: Unbounded Interval Find all extrema of f (x) = 3x4 – 4x3 on [–1, ). First, calculate f (x) = 12x3 – 12x2. Stationary points: Let f (x) = 0 : 12x3 – 12x2 = 0 => 12x2(x – 1) = 0. x = 0 and x = 1, and both are in the domain. Stationary points are: (0, 0) and (1, -1) Singular points: There are no points where f (x) is not defined => no singular points. Endpoints: The domain is [–1, ∞ ), => one endpoint, at x = –1 (-1, 7) To determine which points are Maxima and which are Minima, Either plot the points/sketch the graph or use the 1st derivative test.

Example Continued… cont’d Using the First Derivative Test: List the critical and endpoints in a table, and add additional points as necessary so that each critical point has a noncritical point on either side. Then compute the derivative at each of these points, and draw an arrow to indicate the direction of the graph. The arrows now suggest the shape of the curve of the graph relative maximum at x = –1, neither a maximum nor a minimum at x = 0 relative minimum at x = 1. You can plot the points and sketch the graph to see the relative/absolute extrema. In practice you use a combination of 1st derivative test and sketching

Extreme Value Theorem If f is continuous on a closed interval [a, b], then it will have an absolute maximum and an absolute minimum value on that interval. Each absolute extremum must occur at either an endpoint or a critical point. Therefore, the absolute maximum is the largest value in a table of the values of f at the endpoints and critical points, and the absolute minimum is the smallest value. Example: The function f (x) = 3x – x3 on the interval [0, 2] has one critical point at x = 1. Critical points and endpoints are given in the following table: Absolute maximum value of f on [0, 2] is 2, which occurs at x = 1 Absolute minimum value of f is –2, which occurs at x = 2.

5.2 An Application: Minimizing Average Cost Gymnast Clothing manufactures expensive hockey jerseys for sale to college bookstores in runs of up to 500. Its cost (in dollars) for a run of x hockey jerseys is C(x) = 2,000 + 10x + 0.2x2. How many jerseys should Gymnast produce per run in order to minimize average cost? To Minimize, Let C’(x) = 0 0 < x ≤ 500 0 = -2000/x2 + .2 -.2 = -2000/x2 .2x2 = 2000 X2 = 10000 X = 100 or -100 Thus, one stationary point a x = 100 F’(0) is undefined, but 0 is not in domain Thus, No singular points One end point at x = 500 Minimum average cost is $50 per jersey

5.3 Higher Order Derivatives: Acceleration & Concavity Let s(t) represents the position of a car at time t Velocity : v(t) = s(t) But one rarely drives a car at a constant speed; the velocity itself is changing. The rate at which the velocity is changing is the acceleration. Because the derivative measures the rate of change, acceleration is the derivative of velocity: a(t) = v(t) a(t) = v(t) = (s)(t) = s(t) Example 1: Calculating the 2nd derivative Let f (x) = x3 – x f (x) = 3x2 – 1 f (x) = 6x Example 2 Let t = time in hours since 12 Noon A car’s position at time t is s(t) = t3 + 2t2 miles Find the car’s velocity and acceleration at 2pm Velocity: v(t) = s(t) = 3t2 + 4t mph Acceleration is a(t) = s(t) = v(t) = 6t + 4 mph v(2) = 20 mph a(2) = 16mph

Differential Notation & Acceleration due to Gravity The second derivative of f (x) as f (x). Or…using differential notation: Example: Acceleration Due to Gravity According to the laws of physics, the height of an object near the surface of the earth falling in a vacuum from an initial rest position 100 feet above the ground under the influence of gravity is approximately: S(t) = 100 – 16t2 [feet in t seconds] Find its acceleration: v(t) = s’(t) = -32t ft/s a(t) = s’’(t) = -32 ft/s2 The negative sign indicates the height of the object is decreasing with time. The downward velocity is increasing by 32 ft/s every second. We say that 32 ft/s2 is the acceleration due to gravity, and if we ignore air resistance, All falling bodies near the surface of the earth will fall with this acceleration.

Second Derivative: Concavity & Points of Inflection f (x) > 0 => f is concave up for 1 < x < 3 (Slope increasing) f (x) < 0 => f is concave down for x < 1 & x > 3 (Slope decreasing) F ’’ (x) = 0 or Undefined => Points of inflection (change in concavity) The function below has points of inflection at x = 1 and x = 3.

Example – Inflation U.S. Consumer Price Index (CPI) Values from January 2007 through June 2008. (Inflation occurs when CPI is increasing => I’(t) > 0 - positive slope - ALL below) I(t) = 0.0075t3 – 0.2t2 + 2.2t + 200 (1 ≤ t ≤ 19) t = time in months. t = 1 is Jan 2007 Relative rate of change of the CPI Estimate Inflation rate in January 2008 (t = 13) I(t) = 0.0225t2 – 0.4t + 2.2 Inflation rate = .8025/211.2775 ≈ .00380 (.38% per month) C) When was inflation slowing/speeding up/slowest? Slowing => I’’< 0 (concave down) Speed up =>I’’(t) > 0 (concave up) Slowest when it switches at point of inflection when I’’(t) = 0 .045t - .4 = 0 => t = 8.9 B) Was inflation slowing or speeding up in Jan 2008 If I(t) < 0; the index rises at a slower rate If I(t) > 0; the index rises at a faster rate I(t) = 0.045t – 0.4 I(13) = 0.045(13) – 0.4 = 0.185 I(13) > 0 => inflation speeding up in Jan 2008

The Second Derivative Test for Relative Extrema The graph has two stationary points: relative maximum at x = a (concave down) f’’ (a) < 0) relative minimum at x = b (concave up) f’’ (b) > 0) Example: f (x) = x2 – 2x f (x) = 2x – 2 0 = 2x = 2 => x = 1 Thus, stationary point at x = 1. f (x) = 2 f (1) = 2, which is positive => f has a relative minimum at x = 1. Second Derivative Test for Relative Extrema If f has stationary point at x = c and f (c) exists, Determine the sign of f (c). 1. If f (c) > 0 then f has a relative minimum at x = c. 2. If f (c) < 0 then f has a relative maximum at x = c. 3. If f (c) = 0 the test is inconclusive and you must use the 1st derivative test.

Higher Order Derivatives There is no reason to stop at the second derivative; we could once again take the derivative of the second derivative to obtain the third derivative, f , and we could take the derivative once again to obtain the fourth derivative, written f (4), and then continue to obtain f (5), f (6), and so on (assuming we get a differentiable function at each stage). Higher Order Derivatives We define

5.4 Analyzing Graphs If y = f (x), find the x-intercept(s) by setting y = 0 and solving for x. Find the y-intercept by setting x = 0 and solving for y 2. Extrema: Check Stationary, Singular and EndPoints to Find Maxima and Minima (where f ‘ (x) = 0 or f ‘(x) is undefined) 3. Find points of inflection (where x in domain & f ‘’ (x) = 0 or undef.) 4. Find behavior near points where If f (x) is not defined at x = a. consider and to see how the graph of f behaves as x approaches a: 5. Behavior at infinity: Consider and if appropriate, to see how the graph of f behaves far to the left and right

Example : Analyzing a Graph Analyze the graph of f (x) = Intercepts: x can’t be 0, so No y-intercepts 0 = (Multiply both sides by x2) 0 = x – 1 => x = 1, so x-intercept at (1, 0) 2. Extrema: f ‘ (x) = => X = 2 (Stationary Point) Use 1st derivative test to see if x = 2 is Max or Min f ‘(1) = 1 f ‘(2) = 0 f ‘(3) = -.037 Relative Max Since f ‘(0) is undefined X = 0 might be a singular point. Check: f (0) is also undefined so we do not have a singular point. (If f(0) had been defined we would have had a singular point) Graph So far after #1 & 2

Example : Analyzing a Graph Continued…. Analyze the graph of f (x) = 3. Points of Inflection: f (x) = => X = 3 (Change of concavity) There is a change at x = 3 from concave down to concave up. To see this: Check f ‘’(x) on both sides of 3. (Ex: f ‘’(1) <0 & f ‘’(4) > 0) f (x) is not defined at x = 0, but 0 is not in the domain => no point of inflection at 0. 4. Behavior near points where f(x) is not defined: f(0) is not defined Just to the left of x = 0) Just to the right of x = 0) If x is close to 0 (on either side) the numerator Is close to -1 and the denominator is a very small positive number. Thus, the quotient is a negative number of very large magnitude. Note: Vert. Asymptote at x = 0

Completed Picture of f(x) = Example Continued…. 5. Behavior at infinity: Both 1/x and 1/x2 go to 0 as x goes to or Completed Picture of f(x) = Note f also has a horizontal asymptote at y = 0

5.5 Related Rates Before we learn about related rates, we need a new skill: Implicit Differentiation Consider the equation y5 + y + x = 0 Because we cannot solve for y explicitly in terms of x, we say that the equation y5 + y + x = 0 determines y as an implicit function of x. We cannot find the derivative directly, but we can find it indirectly: Find We must be careful. The derivative with respect To x of y5 is not 5y4 Y is a function of x, so we Must use the chain rule as Follows:

Solving Related Rate Problems Problem: The radius of a circle is increasing at a rate of 10 cm/s. How fast is the area increasing at the instant when the radius has reached 5 cm? Rewrite the problem in terms we know from our study of derivatives:  Two related quantities: the radius of the circle, r, and its area, A. The rate of change of r is 10 cm/s. So, dr/dt = 10 What is the rate of change of A when the radius is 5 cm? What is dA/dt when r = 5 A = πr 2 d (A) = d (πr 2) dt dt Derived Equation Method to find related rate Take the derivative of Both sides of the equation With respect to t. On the Left we get dA/dt. On the Right, r is a function of t. Use the chain rule. Area is increasing at the rate of 314 cm2/s when radius is 5 cm

Falling Ladder Problem Jane is at the top of a 5-foot ladder when it starts to slide down the wall at a rate of 3 feet per minute. Jack is standing on the ground behind her. How fast is the base of the ladder moving when it hits him if Jane is 4 feet from the ground at that instant? Rewrite: What is the rate of change of the distance of the base from the wall when the top of the ladder is 4 feet from the ground? Find when h = 4. h2 + b2 = 25 (4)2 + b2 = 25 b2 = 25 – 16 b2 = 9 => b = 3 Substitute in derived equation h2 + b2 = 25 d/dt ( h2) + d/dt (b2) = d/dt (25) The base of the ladder is sliding away from the wall at 4 ft/min when it hits Jack We can solve for db/dt, but we need “b” first

5.6 Elasticity You manufacture an extremely popular brand of sneakers and wonder what will happen if you increase the selling price. Common sense says demand will drop as you raise the price. Will the drop in demand be enough to cause revenue to fall? (ELASTIC) OR Will the drop in demand be small enough that your revenue rises because of the higher selling price? (INELASTIC) The price elasticity of demand E is the percentage rate of decrease of demand per percentage increase in price. E is given by the formula If E > 1 => Elastic If E < 1 => InElastic If E = 1 => Unit Elasticity (occurs at price that results in the largest revenue)

Elasticity Example1 Suppose that the demand equation is q = 20,000 – 2p where p is the price in dollars. If p = $2,000, then E = 1/4, and demand is inelastic at this price. If p = $8,000, then E = 4, and demand is elastic at this price. If p = $5,000, then E = 1, and the demand has unit elasticity at this price. (thus, this price results in largest revenue) In the above example, if p = $2,000 then the demand would drop by only for every 1% increase in price.

Example2: Price Elasticity of Demand: Dolls Suppose that the demand equation for Bobby Dolls is given by q = 216 – p2 p = price per doll in dollars q = the number of dolls sold per week. Compute the price elasticity of demand when p = $5 and $10 Let p = $5 Let p = $10 When the price is set at $5, the demand is dropping at a rate of 0.26% per 1% increase in the price. Because E < 1, the demand is inelastic at this price, so raising the price will increase revenue. When the price is set at $10, the demand is dropping at a rate of 1.72% per 1% increase in the price. Because E > 1, demand is elastic at this price, so raising the price will decrease revenue; lowering the price will increase revenue.

Example2 Continued… b. Find the price at which the weekly revenue is maximized. What is the maximum weekly revenue? E Set E = 1: Maximum revenue occurs when p = $8.49 Maximum weekly revenue: R = qp => (216 – p2) p (216 – 8.492)(8.49) ≈ $1,222. c. Find the ranges of prices for which the demand is elastic and the range for which the demand is inelastic The demand is elastic when p > $8.49 (the price is too high), and the demand is inelastic when p < $8.49 (the price is too low).