Question 1: What is the extremum and an absolute Question 2: What is - PDF document

Question 1: What is the extremum and an absolute Question 2: What is a critQuestion 3: How do you find the reQuestion 4: How do you find the Many applications in business and economics requidimensions at which the volume is greatest.In this section we’ll find these points by function’s graph as well as the very highest and lowest points on a graph. Question 1: What is the extremum and an absolute relative extremum (the plural ,()cfc there exists an open interval , in the domain of f ()() xfc for all x in , ,()cfc there exists an open interval , in the domain of f ()() xfc , occur at the endpoints. The definition above isbusiness applications. Example 1 Find the Relative Extrema 0,7 Solutionhe highest and lowest points on the tive extrema to illustrate the open 1,146,74,394,39relative maximum. Example 2 Find the Relative Extrema (,8]5,10open interval centered on 5,10 A function’s absolute extremum occurs at the highest or lowest point on a function. The highest point on a function is called the absolute maximum and the lowest point on a Let be a function defined on some interval and c be a number in that interval. ,()cfc ()()fxfc ,()cfc ()()fxfc that are defined on open intervals. Example 3 Find the Absolute Extrema 0,7Find all absolute extrema of Solution0,77,650,7 () 1,140,7Example 4 Find the Absolute Extrema (,8]Find all absolute extrema of Solution Unlike Example 3 this function is not defined on a closed 5,10graph continues downward on the left side (as evidenced by the arrow e is no lowest point. Therefore, () lative extrema or absolute extrema. Figure 1 – A linear function defined on three different intervals. In (a), the function is defined for all real numbers. In (b), the function is defined on the open interval (1, 7). The function in (c) is defined on a closed interval [1,5]. 4.9. If this were the case, these ordered pairs over the open interval. By inspecting the graph closely, we can see that 4.9Since we can always move a little bit closer To guarantee an absolute maximum and absolute 0,1Figure 2 - A discontinuous function defined on a closed interval [0,5].Continuous functions are functions whos 3,4 like 2.993,4 since it is the highest point on the function over the closed 0,5 . The absolute minimum of0,10,5Figure 3 - A continuous function defined on the interval [0,5]. The Extreme Value Theorem gives the conditi Extreme Value Theorem that is continuous on a closed interval is guaranteed to have both an absolute maximum and an absolute minimum. In the rest of this section,e extrema and relative extrema Question 2: What is a critrises as we move from left to right on its Figure 4 – When a function falls and then rises, we get a relative minimum as in (a). In (b), the function rises and then falls resulting in a relative maximum. ft and then drops on the right has a relative maximum. The terms increasing and decreasing Suppose f is defined. ()()fxfx 12 ()()fxfx whenever 12 Figure 5 – In (a), function values on the right are higher than function values on the left so the function is increasing. In (b), function values on the right are smaller than function values on the left so the function is decreasing.lative extrema of a function later in this be a continuous function defined on an open interval containing a number c. The number c is critical value (or critical number) if or is undefined. A critical ,()cfc lue to indicate critical values where the value to indicate critical values where decreasing between the critical values. Test for Determining the In Increasing or Decreasing Suppose a function we are able to compute the derivative of the function at each point in the open interval. ()0 increasing on the interval. ()0 at every x value in the interval, then f is decreasing on the interval. ()0 at every x value in the interval, then f is constant for finding all intervals where a function is Strategy for Tracking the Sign of the Derivative unction and use it to find all of the critical values. Below a number line, label these values. 0 to indicate critical values where the derivative is zero or write und to indicate critical between these values are where we will determine the sign of the derivative. 2. Pick a value in each of the open intervals between the critical values. Substitute these values in () to determine whether the derivative is positive or negative at these values. Label the number line + or – to indicate whether the derivative is positive or negative. 3. The function is increasing over the intervals where ()0()0 Example 5 Find Where a Function is Increasing and Decreasing ()18 xxxquestions below. a. Find the derivative of Solution ()184182436 xxxdxdx ªºªº¬¼¬¼ ()436 xxxb. Use the derivative to find the critical values of Solution are where () is equal to zero or undefined. Since () is a polynomial, it is defined everywhere. All ()0 . Set the derivative equal to zero and solve for x to find these points, Use Sum / Difference Rule and the Product with a Constant Rule for Use the Power Rule for Derivatives 43604904330403030033xxxxxxxxx0,3 c. Find the intervals where is increasing and the intervals where () Solutionhen the derivative of the function is negative. Above each critical number place the (,3)(3,0)(0,3)(3,) . In each of these intervals, the graph is either increasing or decreasing. The critical numbers are where the graph could potentially change from increasing to decreasing of vice versa. Set the derivative equal to zero Factor the greatest common factor, x Factor the difference of squares Set each factor equal to zero Solve each equation for x ()433fxxxx = 0 = 0 -3 = 0 3 To decide where the graph is increas()436 xxx ()433fxxxx For instance, the test point 4x (,3) (4)4443431671 This tells us the function’s graph (,3) we need to test the derivative in (,3) (3,0) (0,3)(3,) . To simplify the testing of the derivative, we need to realize that we only need to know the sign of the derivative, not the actual value. If we substitute the value 4x into the derivative and note the sign of each factor, we can quickly determine the sign of the product of the factors: negativenegativenegative(4)44343   (-)(-)(-) = - ()433fxxxx = 0 -3 = 0 = 0 0 The product of three negative factors is negative so we know the Let’s examine the test point 1x in the interval 3,0examine the signs of the factors in negativenegativepositive(1)11313  Label these signs above the inThe product of two negative numbers and 3,0see exactly where the first derivative is positive and negative. ive is positive in the intervals 3,03,0 . (-)(-)(+) = + (-)(-)(-) = - ()433fxxxx = 0 -3 = 0 = 0 0 (+)(-)(+) = - (+)(+)(+) = + (-)(-)(+) = + (-)(-)(-) = - ()433fxxxx = 0 -3 = 0 = 0 decreasing increasing decreasing increasing (+)(-)(+) = - (+)(+)(+) = + (-)(-)(+) = + (-)(-)(-) = - ()433fxxxx = 0 -3 = 0 = 0 0 With the number line labeled like thgraph is decreasing in the intervals 0,3that the function increases on 3,0ous when using the graph as our Question 3: How do you find the reThe strategy for tracking the si for more than determining evious question allow ng critical point. First Derivative Test is defined at a critical value changes from positive to negative at x c , then a relative maximum occurs at the critical point ,()cfc changes from negative to positive at x c , then a at the critical point ,()cfc does not change sign at x c , then there is no relative extrema at the corresponding critical point. Example 6 Find the Relative Extrema of a Function ()421185fxxxx Solutionfor the derivative so that we can i is ()421185432121810ddddfxxxxdxdxdxdxªºªº¬¼¬¼()124218fxxx Use Sum / Difference Rule and the Product with a Constant Rule. Use the Power Rule for Derivatives and the fact that the derivative of a constant is zero We need to use this derivative to find the critical values. Set the ()124218fxxx, equal to zero to find those values.  1242180627306213021030213  everywhere so the derivative is defined everywhere. Although this derivative could be fact can be solved using the quadratic formula. This strategy woul values as factoring:  1242180424241218212429004230re complicated derivatives, we may Set the derivative equal to zero Factor the greatest common factor from each term Factor the trinomial To find where the product is equal to zero, set each factor equal to zero and solve for the variable Identify a, b, and c for the quadratic bbac Simplify the numerator and denominator 1242180 can also be found by locating the x intercepts on the derivative Like factoring or the quadratic formula, the critical values are located at . All three strategies yield thin mind that a graph will give approximlinear, or quadratic, another method With the critical values in hand, lber line and track the sign of the ()6213fxxx = 0 = 0 3 have a relative maximum at the crthis case, the relative maximum is the original function 111122224Relative Maximum:421185Relative Minimum:343213183522The relative maximum is located at 3,22 (+)(-)(-) = + (+)(+)(-) = - (+)(+)(+) = + increasing decreasing increasing ()6213fxxx = 0 = 0 3 Example 7 Find the Relative Maximum () SolutionDerivatives with x x uxveuve Put these expressions into the Quotient Rule: 1 exe exe Now we can find the critical values by or 1 x . Fractions are undefined where the denominator is equal to zero. The denominator for this fraction is xe and is always positive so there are no x values where the fraction is undefined. The Quotient Rule is duvuuvdxvv Factor from the numerator Reduce the common factor in the numerator and denominator a number line with this critical and the derivative changes from x , the critical point is a relative maximum. The maximum comes from the and is 11e maximum is Example 8 Find the Minimum Average Cost 0.005101000 dollarsCQQQ a. Find the average cost function Solution CQSubstitute the total daily cost function into the numerator of this fraction to yield increasing decreasing = 0 1 0.005101000b. Find the quantity that yielSolution find this relative minimum, we need to take the derivative of the Quotient Rule for Derivatives.0.0051010000.010101uQQvQuQv 0.010100.0051010001QQQQ   simplify the derivative as much as possible: 0.010100.0051010000.0051000QQQQundefined where the denominator is equal to zero. Simplify the numerator by carrying out the multiplication and subtraction Combine like terms Set the numerator equal to 00.005100000.0051000Set the denominator equal to 0The number line for this function C Since the function decreases on the left side of the critical value and 0.005101000 and is calculated as decreasing increasing 0.0051000 = 0 447 0.005200000102000001000Since the total daily cost is in dolla dollars per unit. Figure 6 - The relative minimum for the average cost Question 4: How do you find the minimum. However, under certain conditions a function will automatically have absolute These extrema will occur at the critical values or at the end points on the closed interval. absolute maximum and the lowest value is the absolute minimum. Strategy for Finding Absolute Extrema a on a continuous function defined over a closed interval, 1. Find all critical values for the function interval. 2. Evaluate each critical value in the function f the closed interval in the function 4. The largest function value from steps 2 and 3 is the steps 2 and 3 is the absolute minimum. endpoints. However, the absolute extrema oExample 9 Find the Absolute Extrema of a Function ()8 xxx 1,6Solution This function is a polynomial so it is continuous not only on the closed interval 1,6an located the critical values of this ()8823164 xxxdxdx ªºªº¬¼¬¼ is a polynomial and defined everywhere, the only critical values are due to where the derivative is equal to zero. Set () 01640444040  Apply the Sum / Difference Rule and the Product with a Constant Rule for Derivatives Use the Power Rule for Derivatives Multiply the constants in each term Set the derivative equal to zero Factor the greatest common factor, x , from each term Set each factor equal to zero and solve for 1,6 ()8fxxxFunction Value Ordered Pair 1x (1)811 4x (4)844 6x (6)86606,0 The absolute maximum occurs at 6,0 ()8 xxx Example 10 Find the Abso ln on the closed interval 2,10Solution is continuous over 2,10denominator is equal to zero outside of this interval. This means the quotient is continuous over 2,10derivative with the quotient rule. uxvx The derivative is ln11ln x where the denominator is equal to zero. This occurs when However, this value is outside of the interval [2,10] so it can be ignored. 1ln() equal to zero and solve for x Apply the Quotient Rule duvuuvdxvv Simplify each term in the numerator 1ln()0ln()1ln()1 x  x e [2,10] x 2,10 into() Using the function ln() Function Value Ordered Pair 2x ln(2)(2)0.352,0.35 1 x e ln()()0.37,0.37 ln(10)(10)0.23 10,0.23 The absolute maximum occurs at approximately ,0.37 10,0.23 1ln equal to zero Subtract from both sides Divide both sides by - Convert to exponential form Example 11 Find the Abso ()2.3346.25285.221107.80Rtttt(Source Verizon Annual Reports 2004 through 2009) a. Over the period 2004 to 2009, wSolution 4,9. The average annual service ()2.3346.25285.221107.80Rtttt is found ()2.3346.25285.221107.802.33346.252285.22106.9992.50285.22ddddRttttdtdtdtdtªºªº¬¼¬¼equal to zero and solving for the variable:  6.9992.50285.22092.5092.5046.99285.2226.994.89,8.34 4,9 ()2.3346.25285.221107.80Rttttnd the absolute maximum. Function Value Ordered Pair 4t (4)2.33446.254285.2241107.804,557.8 (4.89)2.334.8946.254.89285.224.891107.80 4.89,546.56 (8.34)2.338.3446.258.34285.228.341107.808.34,594.39 t (9)2.33946.259285.2291107.80 9,588.5 The average annual service revenue per customer is highest at or in the year 2008. Set the derivative equal to zero and identify the coefficients Solve the equation using the quadratic bbac b. When the average annual servSolution dollars per customer. However, this is the . To get the monthly service (8.34)594.391212 dollars per month for each customer. b. When the average annual servSolution dollars per customer. However, this is the . To get the monthly service (8.34)594.391212 or 49.53 dollars per month for each customer. 37  6.9992.50285.22092.5092.5046.99285.2226.994.89,8.34 4,9 ()2.3346.25285.221107.80Rttttnd the absolute maximum. Function Value Ordered Pair 4t (4)2.33446.254285.2241107.804,557.8 4.89t (4.89)2.334.8946.254.89285.224.891107.80 4.89,546.56 8.34t (8.34)2.338.3446.258.34285.228.341107.808.34,594.39 9t (9)2.33946.259285.2291107.80 9,588.5 The average annual service revenue per customer is highest at or in the year 2008. Set the derivative equal to zero and identify the coefficients Solve the equation using the quadratic bbac 36 Example 11 Find the Abso ()2.3346.25285.221107.80Rtttt(Source Verizon Annual Reports 2004 through 2009) a. Over the period 2004 to 2009, wSolution 4,9. The average annual service ()2.3346.25285.221107.80Rtttt is found ()2.3346.25285.221107.802.33346.252285.22106.9992.50285.22ddddRttttdtdtdtdtequal to zero and solving for the variable: 35 1ln()0ln()1ln()1 x  x e [2,10] x 2,10 into() f x. Using the function ln() x fx x Function Value Ordered Pair 2x ln(2)(2)0.352,0.35 1 x e ln()()0.37,0.37 10x ln(10)(10)0.23 10,0.23 The absolute maximum occurs at approximately ,0.37 10,0.23 1ln () x equal to zero Subtract from both sides Divide both sides by - Convert to exponential form 34 Example 10 Find the Abso ln() x fx x on the closed interval 2,10Solution 2,10denominator is equal to zero outside of this interval. This means the 2,10derivative with the quotient rule. uxvx The derivative is ln11ln x where the denominator is equal to zero. This occurs when However, this value is outside of the interval [2,10] so it can be ignored. 1ln() x fxx equal to zero and solve for x: Apply the Quotient Rule duvuuvdxvv Simplify each term in the numerator 33 1,6 ()8fxxxFunction Value Ordered Pair 1x (1)811 2031, 4x (4)844 12834, 6x (6)86606,0 The absolute maximum occurs at 6,0 ()8 f xxx 32 endpoints. However, the absolute extrema oExample 9 Find the Absolute Extrema of a Function ()8 f xxx 1,6Solution This function is a polynomial so it is continuous not only on the closed interval 1,6an located the critical values of this ()8823164 f xxxdxdx x  f x is a polynomial and defined everywhere, the only critical values are due to where the derivative is equal to zero. Set () f 01640444040 x  Apply the Sum / Difference Rule and the Product with a Constant Rule for Derivatives Use the Power Rule for Derivatives Multiply the constants in each term Set the derivative equal to zero Factor the greatest common factor, x , from each term Set each factor equal to zero and solve for 31 Question 4: How do you find the minimum. However, under certain conditions a function will automatically have absolute These extrema will occur at the critical values or at the end points on the closed interval. absolute maximum and the lowest value is the absolute minimum. Strategy for Finding Absolute Extrema To find the absolute extrema on a continuous function f defined over a closed interval, 1. Find all critical values for the function interval. 2. Evaluate each critical value in the function f. the closed interval in the function f. 4. The largest function value from steps 2 and 3 is the absolute maximum and the smallest function value from steps 2 and 3 is the absolute minimum. 30 0.005200000102000001000Since the total daily cost is in dolla dollars per unit. Figure 6 - The relative minimum for the average cost 29 Set the numerator equal to 00.005100000.0051000Set the denominator equal to 0The number line for this function C Since the function decreases on the left side of the critical value and 0.005101000 and is calculated as decreasing increasing 0.0051000 = 0 447 0 28 0.005101000b. Find the quantity that yields the minimum average cost. Solution find this relative minimum, we need to take the derivative of the Quotient Rule for Derivatives.0.0051010000.010101uQQvQ 0.010100.0051010001QQQQ   simplify the derivative as much as possible: 0.010100.0051010000.0051000QQQQundefined where the denominator is equal to zero. Simplify the numerator by carrying out the multiplication and subtraction Combine like terms 27 a number line with this critical x , the critical point is a relative maximum. The maximum comes from the () x x gxe and is 11(1)ge maximum is Example 8 Find the Minimum Average Cost 0.005101000 dollarsCQQQ a. Find the average cost function Solution CQCQQ. Substitute the total daily cost function into the numerator of this fraction to yield increasing decreasing 1() x x fxe = 0 1 26 Example 7 Find the Relative Maximum () x x gxe SolutionDerivatives with x x uxveuve Put these expressions into the Quotient Rule: 21() x exe exe Now we can find the critical values by or 1 x . Fractions are undefined where the denominator is equal to zero. The denominator for this fraction is xe and is always positive so there are no x values where the fraction is undefined. The Quotient Rule is duvuuvdxvv Factor from the numerator Reduce the common factor in the numerator and denominator 25 have a relative maximum at the crthis case, the relative maximum is 12x and the relative minimum is located at 3x. To find the ordered pairs for the relative extrema, we need to substitute the critical values into the original function ()fx to find the corresponding y values: 111122224Relative Maximum:421185Relative Minimum:343213183522The relative maximum is located at 3,22 () f x (+)(-)(-) = + (+)(+)(-) = - (+)(+)(+) = + increasing decreasing increasing ()6213fxxx = 0 = 0 3 24 1242180 can also be found by locating the x intercepts on the derivative () f x. Like factoring or the quadratic formula, the critical values are located at . All three strategies yield thin mind that a graph will give approximlinear, or quadratic, another method With the critical values in hand, label them on a number line so that we ber line and track the sign of the ()6213fxxx = 0 = 0 3 23 We need to use this derivative to find the critical values. Set the ()124218fxxx, equal to zero to find those values.  1242180627306213021030213  f x everywhere so the derivative is defined everywhere. Although this derivative could be fact can be solved using the quadratic formula. This strategy woul values as factoring:  1242180424241218212429004230re complicated derivatives, we may Set the derivative equal to zero Factor the greatest common factor from each term Factor the trinomial To find where the product is equal to zero, set each factor equal to zero and solve for the variable Identify a, b, and c for the quadratic bbac Simplify the numerator and denominator 22 First Derivative Test is defined at a critical value x c. If f changes from positive to negative at x c , then a relative maximum occurs at the critical point ,()cfc If f changes from negative to positive at x c , then a at the critical point ,()cfc If f does not change sign at x c , then there is no relative extrema at the corresponding critical point. Example 6 Find the Relative Extrema of a Function Find the location of the relative extrema of the function ()421185fxxxx Solutionfor the derivative so that we can i f x is ()421185432121810ddddfxxxxdxdxdxdx()124218fxxx . Use Sum / Difference Rule and the Product with a Constant Rule. Use the Power Rule for Derivatives and the fact that the derivative of a constant is zero 21 Question 3: How do you find the reThe strategy for tracking the si for more than determining evious question allow When the derivative of a function changes from positive to negative, we know the changes from decreasing to increasing. In this case, the critical point is a relative ng critical point. 20 0,3We can check this number line by examining the graph ofthat the function increases on 3,0ous when using the graph as our approximate values from the graph. The derivative allows us to find the 19 The product of three negative factors is negative so we know the ,3 . Let’s examine the test point 1x in the interval 3,0examine the signs of the factors in negativenegativepositive(1)11313  Label these signs above the inThe product of two negative numbers and 3,0see exactly where the first derivative is positive and negative. ive is positive in the intervals 3,03,0 3, . (-)(-)(+) = + (-)(-)(-) = - ()433fxxxx = 0 -3 = 0 = 0 0 (+)(-)(+) = - (+)(+)(+) = + (-)(-)(+) = + (-)(-)(-) = - ()433fxxxx = 0 -3 = 0 = 0 decreasing increasing decreasing increasing (+)(-)(+) = - (+)(+)(+) = + (-)(-)(+) = + (-)(-)(-) = - ()433fxxxx = 0 -3 = 0 = 0 0 18 To decide where the graph is increas()436 f xxx ()433fxxxx For instance, the test point 4x (,3) (4)4443431671 This tells us the function’s graph (,3) we need to test the derivative in (,3) (3,0) (0,3)(3,) . To simplify the testing of the derivative, we need to realize that we only need to know the sign of the derivative, not the actual value. If we substitute the value 4x into the derivative and note the sign of each factor, we can quickly determine the sign of the product of the factors: negativenegativenegative(4)44343   (-)(-)(-) = - ()433fxxxx = 0 -3 = 0 = 0 0 17 43604904330403030033xxxxxxxxx0,3 c. Find the intervals where f x is increasing and the intervals where () f Solutionhen the derivative of the function is negative. Above each critical number place the (,3)(3,0)(0,3)(3,) . In each of these intervals, the graph is either increasing or decreasing. The critical numbers are where the graph could potentially change from increasing to decreasing of vice versa. Set the derivative equal to zero Factor the greatest common factor, x Factor the difference of squares Set each factor equal to zero Solve each equation for x ()433fxxxx = 0 = 0 -3 = 0 3 16 whether the derivative is positive or negative at these values. Label the number line + or – to indicate whether the derivative is positive or negative. 3. The function is increasing over the intervals where ()0()0 . Example 5 Find Where a Function is Increasing and Decreasing ()18 f xxxquestions below. a. Find the derivative of f Solution ()184182436 f xxxdxdx x  f ()436 f xxxb. Use the derivative to find the critical values of f Solution f x are where () f x is equal to zero or undefined. Since () f x is a polynomial, it is defined everywhere. All ()0 . Set the derivative equal to zero and solve for x to find these points, Use Sum / Difference Rule and the Product with a Constant Rule for Use the Power Rule for Derivatives 15 Test for Determining the In Increasing or Decreasing Suppose a function we are able to compute the derivative of the function at each point in the open interval. ()0 increasing on the interval. ()0 at every x value in the interval, then f is decreasing on the interval. ()0 at every x value in the interval, then f is constant on the interval. for finding all intervals where a function is Strategy for Tracking the Sign of the Derivative unction and use it to find all of the critical values. Below a number line, label these values. Above the number line, write = 0 to indicate critical values where the derivative is zero or write und to indicate critical values where the derivative is undefined. The open intervals e we will determine the sign of the derivative. 2. Pick a value in each of the open intervals between the critical values. Substitute these values in () f x to determine 14 Figure 5 – In (a), function values on the right are higher than function values on the left so the function is increasing. In (b), function values on the right are smaller than function values on the left so the function is decreasing.lative extrema of a function later in this be a continuous function defined on an open interval containing a number c. The number c is critical value (or critical number) if 0fc or fc is undefined. A critical ,()cfc value to indicate critical values where decreasing between the critical values. 13 Question 2: What is a critrises as we move from left to right on its Figure 4 – When a function falls and then rises, we get a relative minimum as in (a). In (b), the function rises and then falls resulting in a relative maximum. A continuous function that rises on the left and then drops on the right has a relative maximum. The terms increasing and decreasing Suppose which a function f is defined. ()()fxfx 12xx. ()()fxfx whenever 12xx. 12 In the rest of this section,e extrema and relative extrema 11 3,4 like 2.993,4 since it is the highest point on the function over the closed 0,5 . The absolute minimum of0,10,5Figure 3 - A continuous function defined on the interval [0,5]. The Extreme Value Theorem gives the conditions under which the absolute extrema are Extreme Value Theorem that is continuous on a closed interval is guaranteed to have both an absolute maximum and an absolute minimum. 10 4.9. If this were the case, these ordered pairs over the open interval. By inspecting the graph closely, we can see that 4.9Since we can always move a little bit closer To guarantee an absolute maximum and absolute 0,1Figure 2 - A discontinuous function defined on a closed interval [0,5].Continuous functions are functions whos 9 lative extrema or absolute extrema. Figure 1 – A linear function defined on three different intervals. In (a), the function is defined for all real numbers. In (b), the function is defined on the open interval (1, 7). The function in (c) is defined on a closed interval [1,5].In Figure 1a, there are no absolute extrema since the graph 8 1,140,7Example 4 Find the Absolute Extrema (,8]Find all absolute extrema of Solution Unlike Example 3 this function is not defined on a closed 5,10graph continues downward on the left side (as evidenced by the arrow e is no lowest point. Therefore, ()dx 7 Example 3 Find the Absolute Extrema 0,7Find all absolute extrema of Solution0,77,650,7 ()px 6 A function’s absolute extremum occurs at the highest or lowest point on a function. The highest point on a function is called the absolute maximum and the lowest point on a function is called the absolute minimum. Let f be a function defined on some interval and c be a number in that interval. ,()cfc ()()fxfc ,()cfc ()()fxfc that are defined on open intervals. 5 Example 2 Find the Relative Extrema (,8]5,10open interval centered on 5,10 ()dx 4 Solutionhe highest and lowest points on the tive extrema to illustrate the open 1,146,74,394,39relative maximum. 3 Question 1: What is the extremum and an absolute The high and low points on a graph are called the extrema of relative extremum (the plural ,()cfc there exists an open interval ,ab in the domain of f ()() f xfc for all x in ,ab. ,()cfc there exists an open interval ,ab in the domain of f ()() f xfc ,ab. occur at the endpoints. The definition above isbusiness applications. Example 1 Find the Relative Extrema 0,7 2 Question 1: What is the extremum and an absolute Question 2: What is a critQuestion 3: How do you find the reQuestion 4: How do you find the Many applications in business and economics requidimensions at which the volume is greatest.In this section we’ll find these points by function’s graph as well as the very highest and lowest points on a graph. 1