# ISM_T11_C04_B - 204 Chapter 4 Applications of Derivatives...

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204 Chapter 4 Applications of Derivatives 50. y ,x x, x w œ ±" ² ! #±# ³! o crit. pt. derivative extremum value x undefined local min x local max œ! \$ œ" ! % 51. y 2x 2, x 1 2x 6, x 1 w œ ±± ² ±´ ³ o crit. pt. derivative extremum value x 1 maximum 5 x 1 undefined local min 1 x 3 maximum 5 œ± ! œ 52. We begin by determining whether f x is defined at x , where f x xx, x xxx w "" " & %# % # \$# ab o œ ±±´ Ÿ " ±' ´) ³" Clearly, f x x if x , and f h . Also, f x x x if x , and ww w # ## Ä! a b œ± ± ²" "´ œ±" œ\$ ±"# ´) lim h f h . Since f is continuous at x , we have that f . Thus, lim h Ä! w w a b " œ±" fx x , x xx , x w # o œ Ÿ " " ) ³ " Note that x when x , and x x when x . ± ± œ! œ±" \$ ±"# ´)œ! œ œ œ#„ #\$ ' \$ # "# „ "# ± % \$ ) "# „% ) ÈÈ È # But , so the critical points occur at x and x . ¸! Þ)%&²" œ#´ ¸\$ Þ"&& \$ \$ È È crit. pt. derivative extremum value x local max 4 x local min ! ¸ \$Þ"&& ! ¸ ±\$Þ!(* 53. (a) No, since f x x , which is undefined at x . w # \$ ±"Î\$ a b # œ # (b) The derivative is defined and nonzero for all x . Also, f and f x for all x . Á# # œ! (c) No, f x need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed interval of the form a, b would have both a maximum value and minimum value on the interval. ÒÓ (d) The answers are the same as (a) and (b) with 2 replaced by a. 54. Note that f x . Therefore, f x . x x, x or x x , x or x x x, x or x x , x or x oo œœ ± ´ * Ÿ ±\$ ! Ÿ ² \$ ±\$ ´ * ² ±\$ ! ² ² \$ ±* ±\$² ²!  \$ \$ ±* ±\$² ²! ³\$ \$ \$ \$ \$ w (a) No, since the left- and right-hand derivatives at x , are and , respectively. * (b) No, since the left- and right-hand derivatives at x , are and , respectively. œ\$ ±") ")

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Section 4.1 Extreme Values of Functions 205 (c) No, since the left- and right-hand derivatives at x , are and , respectively. œ±\$ ") ±") (d) The critical points occur when f x (at x ) and when f x is undefined (at x and x ). The ww ab È œ! œ „ \$ œ „\$ minimum value is at x , at x , and at x ; local maxima occur at and . ! œ\$ ± \$ ß' \$ \$ Š‹ Š ÈÈ 55. (a) The construction cost is C x x x million dollars, where x miles. The following is a b È œ !Þ\$ "' ² ² !Þ# * ± ! Ÿ Ÿ * # a graph of C x . Solving C x gives x miles, but only x miles is a critical point is w !Þ\$ "' ± )& & œ ± !Þ# œ ! œ „ ¸ „ \$Þ&) œ \$Þ&) x x È È # the specified domain. Evaluating the costs at the critical and endpoints gives C \$3 million, C \$2.694 ¸ & È million, and C \$2.955 million. Thereform, to minimize the cost of construction, the pipeline should be placed from the docking facility to point B, 3.58 miles along the shore from point A, and then along the shore from B to the refinery. (b) If the per mile cost of underwater construction is p, then C x p x x and a b È œ "'² ²!Þ#*± # C x gives x , which minimizes the construction cost provided x . The value w \$! Þ) "' ± ²!Þ!% œ± ! Þ # œ ! œ Ÿ * x x c c p # # of p that gives x miles is . Consequently, if the underwater construction costs \$218,864 per mile or less, c œ * !Þ#"))'% then running the pipeline along a straight line directly from the docking facility to the refinery will minimize the cost of construction.
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## This note was uploaded on 09/20/2010 for the course MATHEMATIC 09991051 taught by Professor Dr.maenshadeed during the Fall '10 term at Norwegian Univ. of Science & Technology.

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ISM_T11_C04_B - 204 Chapter 4 Applications of Derivatives...

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