Abe Mirza Graphing f ( ) Steps to graph f ( ) 1. Set f ( ) = 0 and solve for critical values.. Substitute the critical values into f ( ) to find critical points.. Set f ( ) = 0 and solve for critical values. 4. Substitute the critical values into f ( ) to find inflection point(s). 5. Do the sign chart for f ( ), to determine where f ( ) is increasing or decreasing 6. Do the sign chart for f ( ), to determine where f ( ) is Concave up or down 7. Graph f ( ) f ( ), f ( ) 1 ) f ( ) > 0, f ( ) > 0 Increasing at an increasing rate nd Quad ) f ( ) > 0, f ( ) < 0 Increasing at a decreasing rate 4 th Quad ) f ( ) < 0, f ( ) < 0 Decreasing at an increasing rate rd Quad 4) f ( ) < 0, f ( ) > 0 Decreasing at decreasing rate 1 st Quad Concavity: Concavity is usually best defined with a graph. So a function is concave up if it opens up and the function is concave down if Try to match the 1,,,4 with the above graphs 1
Eample. Given f( ) = 8+ 10 1. Find critical points and all intervals the f ( ) is decreasing and increasing. Find inflection point all intervals the f ( ) is concave up and down, also graph f ( ) = = = Critical Values f (4) = 16 + 10 = 6 Critical Points f ( ) 8 0 4 X 4 f ( ) = 8 0 + y = f( ) Down 6 Up f ( ) = > 0, Always Concave up A. Given ( ) = 1 + 1 f 1. Find critical points and all intervals the f ( ) is decreasing and increasing. Find inflection point all intervals the f ( ) is concave up and down, also graph f ( ) f ( ) = 1 = ( 4) = ( )( + ) = 0 =, Critical Values ( ) 1 1 f = + ( ) 17 f =, f () = 15 (,17) (, 15) Critical Points X = + 0 0 + f ( ) 1 y = f( ) 17 15 f ( ) = 1 + 1 ( ) = 1 f ( ) 6 0 0 f = = = f (0) = 1 Inflection Point X 0 f ( ) = 6 0 + y = f( ) Down 1 Up
B. Given 4 f( ) = 4 1. Find critical points and all intervals the f ( ) is decreasing and increasing. Find inflection point all intervals the f ( ) is concave up and down, also graph f ( ) C. Given f( ) = 10 + 7, on the interval 0 4 1. Find critical points and all intervals the f ( ) is decreasing and increasing. Find inflection point all intervals the f ( ) is concave up and down, also graph f ( ). Find absolute min and ma in the given interval D. Given + f( ) = 1, on the interval 1 1. Find critical points and all intervals the f ( ) is decreasing and increasing. Find inflection point all intervals the f ( ) is concave up and down, also graph f ( ). Find absolute min and ma in the given interval E. For what value(s) of will the function 1 f ( ) = 5, on the interval 10 10 be largest? F: Find positive numbers whose product is 5 and whose sum is a minimum. G What is the maimum value of 4, f( ) = 8 on the interval? H A shepherd wishes to build a rectangular fenced area against the side of a barn. He has 60 feet of fencing material, and only needs to use it on three sides of the enclosure, since the wall of the barn will provide the last side. What dimensions should the shepherd choose to maimize the area of the enclosure? I. Adam has 000 yd of fencing material with which to enclose a rectangular portion of a river. If fencing is not required along the river, what are the dimensions of the largest area he can enclose? What is this Area?
Solution E. f ( ) = 4 5 = ( 5)( + 1) = 0 so f'() = 0 at = 5 and = - 1. Now, to classify these points, the second derivative test can be used. f''() = - 4; f''(5) = 6, which is greater than zero, so f (5) is a local minimum. f''(- 1) = - 6,which is less than zero, so f (- 1) is a local maimum. Finally, the actual values of the local maima and the endpoints should be compared to each other to see which of these the absolute maimum is. f (- 10) = - 8 f (- 1) = f (10) = 8 So, the absolute maimum on the interval occurs when = 10. F. Objective: S = + y. The goal is to minimize S. Constraint: y = 5. Substitute constraint into objective: y = 5 / 5 S = + 0 < < 5 S = 1+ 5 S = 1+ = 0 = 5 50 50 Use second derivative to classify: S = S (5) = = > 0 5 S''(5) > 0, so S has a local min at = 5. However, notice that S''() is always positive on the interval (0, ), so S is always concave up on that interval, which means that the local min is also the absolute min. Therefore, 5 and 5 are the positive numbers with the smallest sum whose product is 5. G. f ( ) = 4 16 = 4 ( 4) = 0 This equals zero when equals 0 or - or +. Making use of the sign of the first derivative yields the following chart for the behavior of f: Based on this, the only local maimum occurs at = 0. Now, to see if this is an absolute maimum on the interval, compare the value of f at = 0 to the value at the endpoints. f (- ) = 6 f (0) = - f () = 6 While = 0 is a local maimum, it is not the absolute maimum on this interval. The absolute maimum occurs at both of the endpoints in this case. 4
H. Below is a sketch of the situation: Objective: maimize A = y. Constraint: y + = 60. Substitution into objective: A = (60 - y)(y) A(y) = 60y - y A'(y) = 60-4y A'(y) = 0 at = 90 A''(y) = - 4, so A''(90) < 0 and y = 90 is a local maimum. However, because A''(y) = - 4 for all y, the graph of A(y) is always concave down, so the local maimum is also the absolute maimum. Thus, choosing y = 90ft and = 180 ft will generate the largest area. I 5