A relative maximum for a function occurs when O dy = 0 and day 0 dx da2The graph of f (a) = x2 +1 is concave down through the interval O (-0o, - V/3) u(0, V/ 3) O (-oo, - 1) u(0, V/3) O (-oo, - 1) u(1, 00) O (-oo, - 1/ 3) u(1, 00)If the function f (3:) = x3 +ma:2 has an inflection point at :1: = 2 the value of "m" '3 C12 03 .4; The total number of relative maximum and minimum points of the function whose derivative, for all x, is given by f'(a:) = a:(ar: 3)3(ac + 1)4 03 0 None of these 3 Find all of the critical numbers of f(x) = (9 ac2) 5 O 3, 0, and 3 O 3 only 0 3 and 3 only C] 0 and 3 only C] 0 only 3x 3 Select the true statement about the rational function g ( a ) = 2 +3 O It passes through ( -1, 0) O there is a horizontal asymptote at y = 0 O It has two vertical asymptotes O It demonstrates odd symmetryThe absolute maximum and minimum values of the function f ( ac) = ac4 - 4x2 + 2 on the interval [-3, 2] are O Maximum: (0, 2) Minimum: CON - 2 O Maximum: (0, 2) U ( #2, 2) Minimum: ( # V 2, -2) O Maximum: (-3, 47) Minimum: F 2 3 - 2 O Maximum: (-3, 47) Minimum: ( + 1 2, -2)Which of the following correctly describes the function f (x) _ 3x- -12x (ac2 -16) O Domain: x * -4 VA: x = -4 HA: y = 0 O Domain: x * -4 VA: x = 4 HA: y = 3 O Domain: x * 14 VA: x = 14 HA: y = 0 O Domain: x # 14 VA: x = -4 HA: y = 3\fMath 31 Formula Sheet PreCalculus Applications of Derivatives 43 + BB = (4+ B)42 - AB+ B2 ) $2 - 51 V2 - V1 V avg a avg 43 - B' = (4-B)42 + AB+ B2) 12 - 1 1 12 - 1 1 Triangle: Rectangle: If ax +bx+c=0, then x=-bivb' -4ac =bh A =L . W 2a 2 P = sum of sides P = 2L + 2w m = 2 - V1 y - y1 =m (x-x, ) Circle: Closed Box: (Rect. Prism) X2 - X1 A = m-2 V =L . w. h D= V(x2 -x, ) +(12 - V1) C = 2ar SA = 2L . w+ 2w.h+ 2L .h Sphere: (ball) Closed Cylinder: (can) V = arch Limits SA = 41 1-2 SA = 2xr2 + 2xrh lim f(x) - fla) lim flath)- fla) h-0 Right Angle Cone: x - a h V = arch a(r" -1) if r#1 r-1 SA = 2mrs where s = Vh2+12 S .. a if -1 0 sin x cos x - 1 lim = 0 1- cos x lim [ v(t ) at = s(1) + so Ja(t) at = v(1) + vo