Chapter 5 Review

Chapter 5: Using the derivative



• SECTION 5.1: Local maxima and minima
  
  
  1. Suggested Problems: #2, 8, 10, 16, 17.
     
  2. Recall that the derivative of a function f(x) tells us when the function is increasing and decreasing:
     
     1. If f(x) > 0 on an interval, then f(x) is increasing on that interval.
        
     2. If f(x) < 0 on an interval, then f(x) is decreasing on that interval.
        
  3. If x₀ is a point in the domain of f(x), then
     
     1. f(x) is said to have a local maximum at x₀ if f(x) f(x₀) for all values of x near x₀.
        
     2. f(x) is said to have a local minimum at x₀ if f(x) f(x₀) for all values of x near x₀.
        
  4. Let f(x) be a function. A point x₀ in the domain of f(x) for which either f(x₀) = 0 or f(x₀) does not exist is called a critical point of f(x). In this case, the point (x₀, f(x₀)) is also referred to as a critical point. If x₀ is a critical point of f(x), then the value f(x₀) is called a critical value of f(x).
     
  5. If f(x) is continuous on an interval, then any point at which it has a local maximum or minimum must also be a critical point of f(x).
     
  6. Suppose that x₀ is a critical point of the function f(x).
     
     1. If f(x) changes from increasing to decreasing at x₀, then f(x) has a local maximum at x₀.
        
     2. If f(x) changes from decreasing to increasing at x₀, then f(x) has a local minimum at x₀.
        
     3. If f(x) does not change from increasing to decreasing, nor from decreasing to increasing, at x₀, then f(x) has neither a local maximum nor a local minimum at x₀.
        
  7. Suppose that x₀ is a critical point of the function f(x).
     
     1. If f(x) changes from positive to negative at x₀, then f(x) has a local maximum at x₀.
        
     2. If f(x) changes from negative to positive at x₀, then f(x) has a local minimum at x₀.
        
     3. If f(x) does not change sign at x₀, then f(x) has neither a local maximum nor a local minimum at x₀.
        
  
  
  
• SECTION 5.2: Inflection points
  
  
  1. Suggested Problems: #2, 7, 8, 9, 15, 17, 26, 28.
     
  2. A point at which the graph of a function y = f(x) changes concavity is called an inflection point of f(x).
     
  3. Inflection points occur only where either f(x₀) = 0 or f(x₀) does not exist. But in order for f(x) to have a point of inflection at x₀, the function f(x) must be continuous across x₀, and the sign of f(x) must change across x₀. If so, then we would say that x₀ is an inflection point of f(x), or even that (x₀, f(x₀)) is an inflection point of f(x).
     
  4. To find the inflection points of y = f(x), find all of the points in the domain of f(x) for which either f(x) = 0 or f(x) does not exist. Make sure that f(x) is continuous at each of these points. If so, and if f(x) changes sign across each point, then each point is a point of inflection.
     
  5. The inflection points of f(x) are the points at which f(x) has local maximum and minimum values.
     
  6. Be able to draw the graph of a function f(x) given information about its derivative f(x) and its second derivative f(x).
     
  
  
  
• SECTION 5.3: Optimization: profit and revenue
  
  
  1. Suggested Problems: #11, 12, 13, 16,17, 20.
     
  2. If x₀ is a point in the domain of f(x), then
     
     1. f(x) is said to have a global maximum at x₀ if f(x) f(x₀) for all values of x.
        
     2. f(x) is said to have a global minimum at x₀ if f(x) f(x₀) for all values of x.
        
  3. Let f(x) be a continuous function defined on an interval of the form a x b (an interval that contains its endpoints). Then f must have both a global maximum and a global minimum value on this interval. If f(x) is not continuous on this interval, or if the interval does not contain its endpoints, then f may have neither a global maximum nor a global minimum value on this interval.
     
  4. To find the global maximum and global minimum values of a continuous function f(x) on an interval of the form a x b, first find the critical points of the function f that are contained within a x b, and then evaluate f(x) at each of these critical points and at the two endpoints. The largest of these (the largest y-value) is the global maximum, and the smallest of these (the smallest y-value) is the global minimum.
     
  5. To find the global maximum and global minimum values of a continuous function f(x) on an interval that does not include both of its endpoints, first find the critical points of the function f that are contained within the interval, and then evaluate f(x) at each of these critical points. Then sketch a graph of the function. If the function has a largest y-value on this interval, then it is the global maximum. If the function has a smallest y-value on this interval, then it is the global minimum.
     
  6. Often, but not always, profit is maximized whenever marginal cost is the same as marginal revenue. However, this can also occur when profit is minimized.
     
  
  
  
• SECTION 5.4: Average cost
  
  
  1. Suggested Problems: #3, 4, 5, 7, 10, 11, 12.
     
  2. If C(q) is the cost of producing q units of a quantity, then the average cost, a(q), of producing q units of that quantity is given by
     a(q) = C(q)/q.
  3. Geometrically, the average cost of producing q units of a quantity is the same as the slope of the line that passes from the point (0, 0) to the point (q, C(q)) on the cost function curve:
     a(q) = (C(q) - 0)/(q - 0).
  4. Often, but not always, the critical points of average cost occur when marginal cost equals average cost.
     
  5. The derivative of a(q) yields
     a(q) = (C(q) - a(q))/q. We can then conclude:
     
     1. The values of q that make a(q) = 0 are the same values of q for which C(q) = a(q) (marginal cost is the same as average cost).
        
     2. If marginal cost is less than average cost, then a(q) < 0, so that average cost is reduced by increasing production.
        
     3. If marginal cost is greater than average cost, then a(q) > 0, so that average cost is increased by increasing production.
        
  
  
  
• SECTION 5.5: Elasticity of demand
  
  
  1. Suggested Problems: #10, 11, 12, 13.
     
  2. Given a product, the elasticity of demand with respect to price is given by the quantity
     E = | (p/q) (dq/dp) |, which is the absolute value of the ratio of the fractional change in demand to fractional change in price.
     
  3. The elasticity of demand E must necessarily be nonnegative, and
     
     1. if E > 1, then demand is said to be elastic and revenue is increased by lowering the price;
        
     2. if E < 1, then demand is said to be inelastic and revenue is increased by raising the price;
        
     3. if E = 1, then p and q correspond to a critical point of the revenue function.
        
  4. If elasticity is large, then a change in price will cause a large change in the number of sales.
     
  5. The demand for a luxury item tends to be elastic, whereas the demand for an item that is a necessity tends to be inelastic.
     
  
  
  
• SECTION 5.6: Logistic growth
  
  
  1. Suggested Problems: #3, 4, 8.
     
  2. The usual model of population growth gives population size as an exponential function of time:
     P(t) = P₀ · a^t, where P₀ = P(0), and a is some positive constant with a = P(t + 1) / P(t). However, this model is not practical in an enclosed system since it implies that if a > 1, then the population becomes unbounded in size as t increases indefinitely. This cannot happen in an enclosed system since resources are limited, and as the population increases, the resources will be consumed faster. Thus there must be a limit on how large the population can become.
     
  3. A logistic function is a function of the form
     f(t) = L / (1 + C e^-kt), where each of L, C, and k are positive constants.
     
  4. The logistic function
     f(t) = L / (1 + C e^-kt), has the following properties:
     
     1. the value of the function at t = 0 is given by f(0) = L / (1 + C);
        
     2. the function is approximately an exponential function with growth rate k for small values of t, given by y = f(0) · e^kt, where f(0) = L / (1 + C);
        
     3. the function is increasing for all values of t;
        
     4. as the value of t approaches -, the value of the function approaches 0;
        
     5. as the value of t approaches , the value of the function approaches L;
        
     6. the function has one inflection point, given by (ln(C) / k, L / 2), and it occurs at the height L / 2, midway between 0 and L.
        
     7. the function is concave up for t < ln(C) / k, and concave down for t > ln(C) / k.
        
  5. Because of the above properties of the logistic function, it is often used to model population size through time:
     P(t) = L / (1 + C e^-kt). The initial size of the population is given by P(0) = L / (1 + C), and the limiting size of the population, L, is called the carrying capacity of the population. The point of diminishing returns---the point where population growth is greatest (the inflection point)---occurs when t = ln(C) / k.
     
  6. Given the size of a population P(t) at time t, expressed as a logistic function of t
     P(t) = L / (1 + C e^-kt), be able to find
     1. the initial size of the population P(0) = L / (1 + C);
        
     2. the carrying capacity of the population L;
        
     3. the point of diminishing returns, t = ln(C) / k.
        
  
  
  

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Last modified: Sat Jun 22 2002