Question

Antiderivative help
f ''(t) = 9/t^(1/2), f(4) = 32, f '(4) = 9
find f(t)

Answer 1

\[f''(t)=\frac{ 9 }{ \sqrt{t} } \therefore f''(t)=9t ^{-1/2}\]

Answer 2

f''(t) = 9 / t^(1/2)
= 9 * t^(-1/2)
can you integrate it from here?

Answer 3

the problem im having is with the constants. i am getting \[12t ^{2/3}+t/4-65 \] after integrating twice. if it were not for the -65 i would be correct

Answer 4

what did you get before that step?

Answer 5

\[18\sqrt{t}+t/4\]

Answer 6

okay thats good, factor out a t^1/2

Answer 7

omg, i had the latex written and accidentally deleted

Answer 8

:( , hold on i'll rewrite it >.>

Answer 9

Thats the worst

Answer 10

how do you factor t^1/2 out of t/4

Answer 11

also it should be \[18\sqrt{t}+t/4+c\]

Answer 12

since\[f'(x)=18\sqrt{t}+t/4+c\] and f'(4)=9 you should get c= 1/4

Answer 13

alright, so i got a answer : \[12t^\frac{ 3 }{ 2 }-27t+92...\]

Answer 14

i believe

Answer 15

t^1/2 [ 18+( (t^1/2)/4) ] would be how you factor it

Answer 16

alanMW web assign says your wrong too. :( the amount of times ive worked this out i swear i should have gotten it right

Answer 17

ive given up. i ran out of guesses. :( thanks anyways though everyone! if you find an answer let me know!

Answer 18

actually i got

\[12t^\frac{ 3 }{ 2 }-27t+44\]

Answer 19

i am about to close this and ask another question about antiderivatives. @AlanMW i cant check it or i would tell you if you are right.

Answer 20

im thinking thats it

Answer 21

@alanmw if you take the derivative of it twice it comes out right, but my answer did as well. ill give you a medal though

Answer 22

It can be factored like this (sorry i forgot to push send) \[18\sqrt{t} + \frac{ t }{ 4 } ------------>>>>>>> 2 t ^{\frac{ 1 }{ 2 }} [ 9 + \frac{ ^{t ^{\frac{ 1 }{ 2 }}} }{ 8 }]\]