Question

https://gyazo.com/39fe74c2b3cf6a155a251ae62440a283
5/ For 0 ≤ x ≤ 6, express g(x) in terms of x. Do not include +C in your final answer.
please help me with number 5

Answer 1

@pink33

Answer 2

it won't work from my laptop

Answer 3

@pink33
what do i have to do?

Answer 4

@freckles

Answer 5

@IrishBoy123 @zepdrix

Answer 6

@SithsAndGiggles

Answer 7

@SithsAndGiggles

Answer 8

@bibby

Answer 9

for 0 ≤ x ≤ 6 ...
\[g(x) = \int\limits_{-2}^{x} (-t+3) dt\]

Answer 10

how can i integrate that?

Answer 11

you would integrate as you normally wold and when the time comes, plug in x for t

Answer 12

i got -1/2 (x-8)(x+2)
what do i have to do next please?

Answer 13

@jigglypuff314

Answer 14

nice job! I didn't think you needed to factor it, but that should be your answer :)

Answer 15

can you please help me with another one, just one more i will be done :D

Answer 16

sure :)

Answer 17

i'm stuck with number 2

Answer 18

i used MVT but my teacher said it wrong because i cannot use the converse of it

Answer 19

I got that
acceleration of water from (0-2) = 0.275
from (2-3) = 0.3
from (3-7) = 0.3
from (7-8) = 0.26

I can say that R''(t) = 0 can be true because the acceleration decreases from positive to negative (3-8)
but I can't say that R'(t) = 0 because the acceleration remains positive throughout...

Answer 20

@jigglypuff314
thank you so much :D

Answer 21

do you know how I got those numbers? :)

Answer 22

yes i do :D

Answer 23

good ^_^

Answer 24

but i have one question, what happen to R(x) when R'(x) = 0?

Answer 25

R'(t) = 0 refers to the moment when the acceleration of the water equals 0
this is usually apparent when goes from increasing to decreasing, or decreasing to increasing

Answer 26

*apparent when R(t) goes from

Answer 27

because that means that R'(t) would be going from positive to negative, or negative to positive (respectively to the scenario above) and pass through 0 on the way there