Question

Let f(x) = 3r" +44.23 + 204r? + 288. - 3. Calculate the derivative f'(x) = Calculate the second derivative f''(x) Note intervals are entered in the format (-00,5)U(7,00) (these are two infinite intervals). Enter "DNE" if the interval is empty. On what interval(s) is f increasing? Increasing: On what interval(s) is f decreasing? Decreasing: On what interval(s) is f concave downward? Concave Down: On what interval(s) is f concave upward? Concave Up: What is the limit as x approaches infinity? Limit is: What is the limit as x approaches negative infinity? Limit is: Sketch the graph and submit with your work.

Answer 1

3 2 Given f«) = 324 +44'+ 2043 +2883–3. differentiate f(x) with respect to a. f'Cod (3x++4422+ 204** +2888-3) dx = 3(42)+44(372) +204 (2x)+286 -(0) $6x) = 12x*+1323+ + 408x +288 differentiate with respect to a f'() = 1213*5+ 132(2x) +408 $'cse) = 360? +2642+408 Set f'(x)=0, to get the critical points 121+1322? + 408x+288 = 0 (x+1)(+) (x+6) = 0 2= -6, -4,-1, are coitical values of ze for 4 46-6) = 3 (4-6)*+44(-6)+204(-6342886-6-3 3 x=-by =-3 for se= -4, f(-4) = 364)*+44(+4)7 2046-437288(4)-3 = 61. for x=-1, $(+1) = 3(-104+44(+1) +2046-1372884)-3 =-128 The Coitical points are (-6,-3), (-4,61) and (-1,-128)

|-6<3 <-41 213-5 interval | x-value f(x) = 192271322+406x+288 Comment -alal-6 x=-7 12 6-7)+1326-2)5 4081-74299 de creating - 216<o 12(-5)² +132(-5)²440845) increasing = 7288-48>o |-4<-1 2 = -2 12(-2)3+1326-2)*7408(-2) decreasing +288 = -96 co -X x=0 288 >o Increasing From the table Intervals of increase is (-6,-4) (-1,0) Intervals of decrease is (-0,-6) u (-4,- To get inflection Points, ser f'(a)=0 36x+264x+40820 2 = -22+ 6 -11-19 = -50120, x=-11+vig 3 3 = -2.214

between x < x <- 5.120 The value Second derivative is Positive and also -9.32loo The value of second derivative is Positive Concave upward on (-6, -5,130) 06-9.214,00 Between - 5.1202XL-2.214 The value of second ） derivative is are Concave down ward on (-5.120,- 2.214)