Question 4: How do you find the absolute etrema of a function? The absolute etrema of a function is the highest or lowest point over which a function is defined. In general, a function may or may not have an absolute maimum or absolute minimum. However, under certain conditions a function will automatically have absolute etrema. The Etreme Value Theorem guarantees that a continuous function defined over a closed interval will have both an absolute maimum and an absolute minimum. These etrema will occur at the critical values or at the end points on the closed interval. To find the absolute etrema from these possibilities, we must determine which of these values yields the highest and lowest values of the function. This is done by testing each critical value and the endpoints in the function. The highest value of the function is the absolute maimum and the lowest value is the absolute minimum. Strategy for Finding Absolute Etrema To find the absolute etrema on a continuous function f defined over a closed interval,. Find all critical values for the function f on the open interval.. Evaluate each critical value in the function f.. Evaluate each endpoint of the closed interval in the function f. 4. The largest function value from steps and is the absolute maimum and the smallest function value from steps and is the absolute minimum. 0
Notice that the absolute etrema are the function values, not the critical values or endpoints. However, the absolute etrema occur at points on the graph given by an ordered pair. Eample 9 Find the Absolute Etrema of a Function Find the absolute maimum and absolute minimum of the function on the closed interval, 6. f ( ) 8 4 Solution This function is a polynomial so it is continuous not only on the closed interval, 6, but everywhere. The absolute etrema of a continous function over a closed interval will occur at a critical value in the interval or at the endpoints. We can located the critical values of this function from the derivative, d 4 d f ( ) 8 d d 4 8 Apply the Sum / Difference Rule and the Product with a Constant Rule for Derivatives Use the Power Rule for Derivatives 6 4 Multiply the constants in each term Since f ( ) is a polynomial and defined everywhere, the only critical values are due to where the derivative is equal to zero. Set f ( ) equal to zero and solve for : 064 04 4 40 40 Set the derivative equal to zero Factor the greatest common factor, 4, from each term Set each factor equal to zero and solve for 0 4
One of these critical values, 0 is not in the interval, 6 so we can ignore it. The other critical point at 4 and endpoints at and 6 are substituted into points on the graph. f( ) 8 in order to find the highest and lowest 4 Values Function Value Ordered Pair 0 0 f, 4 () 8 8 8 f 4, 4 4 (4) 84 4 f 6,0 4 6 (6) 86 6 0 f ( ) 8 4 The absolute maimum occurs at 4, since it has the largest y value. The absolute minimum occurs at 6,0 since it has the smallest y value. 8
Eample 0 Find the Absolute Etrema of a Function Find the absolute maimum and absolute minimum of the function ln f( ) on the closed interval,0. Solution The natural logarithm is continuous over,0 and the denominator is equal to zero outside of this interval. This means the quotient is continuous over,0. Therefore the absolute etrema are located at the critical values or the endpoints of the interval. To find the critical values, calculate the derivative with the quotient rule. The numerator, denominator and their derivatives are u ln v u v The derivative is f ln ln d u vu uv Apply the Quotient Rule d v v Simplify each term in the numerator The critical values of a function are where the derivative is equal to zero or undefined. For a fraction like this one, the derivative is undefined where the denominator is equal to zero. This occurs when 0. However, this value is outside of the interval [,0] so it can be ignored. To find where the derivative is equal to zero, set the numerator of ln( ) f( ) equal to zero and solve for :
ln( ) 0 ln( ) ln( ) e Set ln( ) equal to zero Subtract from both sides Divide both sides by - Convert to eponential form This critical value is at the interval [,0]. e. This value is approimately.7 and is in To find the absolute etrema, we need to substitute the critical value at e and the endpoints of the interval at,0 into f ( ). ln( ) Using the function f( ), we get the location of each ordered pair at the critical number and endpoints. Values Function Value Ordered Pair ln() () 0.5 f,0.5 e ln( e ) ( ) 0.7 e,0.7 f e e 0 ln(0) (0) 0. 0 f 0,0. The absolute maimum occurs at approimately e,0.7 and the absolute minimum occurs at approimately 0,0.. 4
Eample Find the Absolute Etrema of a Function Verizon Wireless charges each customer a monthly charge for service on its wireless network. This charge is recorded as service revenue in corporate reports. The average annual service revenue per customer (in dollars) at Verizon Wireless from 004 to 009 can be modeled by the function Rt t t t ( ). 46.5 85. 07.80 where t is the number of years since 000. (Source Verizon Annual Reports 004 through 009) a. Over the period 004 to 009, when was the the average annual service revenue per customer highest? Solution Since this function is defined from 004 to 009, the variable t is defined on the closed interval 4,9. The average annual service revenue per customer is highest at the absolute maimum on this interval. The derivative of Rt ( ).t 46.5t 85.t 07.80 is found using the basic rules for taking derivatives and the Power Rule for Derivatives: d d d d R t t t t dt dt dt dt t t ( ). 46.5 85. 07.80. 46.5 85. 0 t t 6.99 9.50 85. Critical values for this polynomial are found by setting this derivative equal to zero and solving for the variable: 5
6.99 t 9.50 t85. 0 a b c t 6.99 9.50 9.50 4 6.99 85. 4.89,8.4 Set the derivative equal to zero and identify the coefficients Solve the equation using the quadratic b b 4ac formula t and simplify a Both of these critical values are in the interval 4,9. Evaluate Rt ( ).t 46.5t 85.t 07.80 at the two critical values and the endpoints to find the absolute maimum. Values Function Value Ordered Pair t 4 R(4). 4 46.5 4 85. 4 07.80 557.80 4,557.8 t 4.89 R(4.89). 4.89 46.5 4.89 85. 4.89 07.80 546.56 4.89,546.56 t 8.4 R(8.4). 8.4 46.5 8.4 85. 8.4 07.80 594.9 8.4,594.9 t 9 R(9). 9 46.5 9 85. 9 07.80 588.5 9,588.5 The average annual service revenue per customer is highest at t 8.4 or in the year 008. 6
b. When the average annual service revenue per customer was highest, how much was each customer paying each month? Solution The average annual service revenue per customer was highest at t 8.4 at a value of 594.9 dollars per customer. However, this is the annual service revenue per customer. To get the monthly service revenue per customer at this time, divide this value by to give R(8.4) 594.9 49.5 or 49.5 dollars per month for each customer. 7