Question

Calc problem that is KILLING me...

Answer 1

Local Min @ S and 3S
Local Max @ 2S and 4S
Absolute Max @ 4S

Concave Down @
(3/2 S, 5/2 S) U (7/2 S, 9/2 S)

Answer 2

yah that ^

Answer 3

it isnt taking that answer, for some reason. webassign is saying to 'check my syntax'

??

Answer 4

This is what i mean:

Answer 5

try replacing s with 4 and where ever an s is next to a number multiply ie 3s would be 4*3 = 12

Answer 6

i tried that, still isnt working >:(

Answer 7

try (16,3)

Answer 8

for absolute max?

Answer 9

yes

Answer 10

nope still saying check your syntax

Answer 11

?!?

Answer 12

16?

Answer 13

tried 16

Answer 14

still doesnt work

Answer 15

umm have you tried blowing it up with dynamite? XD im sorry if its none of those i honestly dont know you're going to have to wait for someone else to answer, btw be sure to re-ask your question seeing this one has been answered

Answer 16

The questions are about g(x), not f(x). Notice that when f(x) is positive, g(x) is increasing because you are adding area. If f(x) is negative, then g(x) is decreasing. The points where there is a local extrema are the points where f(x) changes from positive to negative (or negative to positive), because this are the points when g(x) stops increasing and starts decreasing (or vice-versa), the points where the derivative is zero. The points where g(x) stops increasing and starts decreasing are maxima and the points where it stops decreasing and starts increasing are minima. This way, we get \[\frac{s}{2} = 2\]and\[\frac{5s}{2} = 10\]as local maxima and\[\frac{3s}{2} = 6\]and\[\frac{7s}{2} = 14\]as local minima.

The absolute maximum is the place where you have added the most area. From 0 to s/2 you add a little area, then subtract from s/2 to 3s/2, add more from 3s/2 to 5s/2, subtract more from 5s/2 to 7s/2 and add from 7s/2 to 9s/2. The area you add from 3s/2 to 5s/2 is more than the area you subtract from s/2 to 3s/2, so\[g(5s/2) > g(s/2)\]and the same happens for the next intervals, so\[g(9s/2) > g(5s/2) > g(s/2)\]and the global maximum is at\[\frac{9s}{2} = 18.\]

Finally, the point is concave downward when the derivative of f is decreasing, because you are adding less and less area. This happens on\[(s/4, s) \cup(2s,3s)\cup(4s,9s/2) = (1,4)\cup(8,12)\cup(16,18).\]

Answer 17

now thats a pro answer lmao

Answer 18

Good god, well done mate! :D that worked perfectly...I owe you one. more than one actualy. is there a way to give out multiple medals?

Answer 19

I have my eye on you, krebante. and DevinBlade, thanks for ur help mate! :)

Answer 20

man that was all him o_o