Question

find the values of a,b and c that make the function \[f(r) = ar^{2} + br + c; \] \[r<0 , 2\sin^{2}(r) + 3\cos^2(r)\] \[r \ge 0\] twice differentiable at r=0.

Answer 1

hard question... have no idea about how to solve it

Answer 2

what is the second function?

Answer 3

thats what is given in the question :D

Answer 4

err its ODTU universities calculus 1 final exam question

Answer 5

do they represent value of r

Answer 6

satellite, let me repost question?

Answer 7

well a first though is that if it is going to be differentiable at 0 at least it better be continuous there. and further since
\[f'_-(r)=2ar+b\] and
\[f'_-(0)=0\] it should also be the case that
\[f'_+(0)=4\sin(r)\cos(r)-6\cos(r)\sin(r)=0\]

Answer 8

so i am thinking that the first thing we know is that
\[c=3\]

Answer 9

hmm

Answer 10

it is not obvious to me how we get a and b because both derivatives are 0 at 0

Answer 11

hmm i get this question from ODTU's final calculus exam

Answer 12

maybe from the second the second derivative? lets see

Answer 13

mybe, lets try

Answer 14

lets call some other people for help? the more brains focus in one question, the easier we get the result

Answer 15

i wrote this out in another window! let me repost

Answer 16

hi korcan, lets see from
f '(r)=4sin r cos r -6cos r sinr
=-2sin r cos r=0, r=720 degrees and
f ''(r)=-2[-sin r sin r+cos r cos r]
=2[(cos r)^2 -(sin)^2]=0, r=45 degrees

Answer 17

lets build up the puzzle little by little...lol

Answer 18

err mark o. we satellite already solved it :D (reposted the question )

Answer 19

can you have a look at my other question? its at left :D starts with "given that..."

Answer 20

ah ok ,,you did? where at

Answer 21

http://openstudy.com/groups/mathematics/updates/4e6b8b830b8b4a2b95db07d2#/groups/mathematics/updates/4e6b811e0b8b4a2b95dae801

Answer 22

and there is new question i asked : http://openstudy.com/groups/mathematics/updates/4e6b8b830b8b4a2b95db07d2#/groups/mathematics/updates/4e6b8b830b8b4a2b95db07d2