Question

let f be the function satisfying f'(x)= xsqrtf(x), for all real numbers , where f(3)=25. Find f"(3). i need help with this :/

Answer 1

\[\Large f \ ' (x) = x*\sqrt{f(x)}\] right?

Answer 2

yes @jim_thompson5910

Answer 3

what is the value of f ' (3) ? Are you able to compute this?

Answer 4

3*sqrt f(3)

Answer 5

\[\Large f \ ' (x) = x*\sqrt{f(x)}\]
\[\Large f \ ' (3) = 3*\sqrt{f(3)}\]
yep, keep going

Answer 6

since we're given f(3) = 25, you can replace f(3) with 25 and simplify further

Answer 7

okay so f'(3)=15?

Answer 8

correct

Answer 9

that will be used when computing f '' (3)

Answer 10

okay what is the next step

Answer 11

\[\Large f \ ' (x) = x*\sqrt{f(x)}\]
\[\Large f \ '' (x) = \sqrt{f(x)}+x*\frac{1}{2\sqrt{f(x)}}*f \ ' (x)\]
\[\Large f \ '' (3) = \sqrt{f(3)}+3*\frac{1}{2\sqrt{f(3)}}*f \ ' (3)\]
\[\Large f \ '' (3) = ???\]
I used the product rule to derive x*sqrt(f(x)) on step 2

Answer 12

also, the chain rule is being applied on step 2

Answer 13

so i got 5+3(1/10)(15)

Answer 14

looks good

what do you get when you simplify that into one fraction?

Answer 15

19/2

Answer 16

I'm getting the same thing

Answer 17

okay well thank u :)

Answer 18

sure thing

Answer 19

are you sure 5+ (3/10)*15 =5+9/2= 9(1/2)