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April 18, 2014

April 18, 2014

Posted by **Kay** on Thursday, June 23, 2011 at 8:25pm.

Find f.

f ''(x) = 3e^x + 3sin(x)

f(0) = 0

f(π) = 0

My work:

f ''(x) = 3e^x + 3sin(t)

f'(x) = 3e^x - 3cos(t) + C

f(x) = 3e^x -3sin(t) + Cx + D

0=f(0)= 3e^0 - 3sin(0) + C(0) + D

D=-3

0=f(π)= 3e^π - 3sin(π) + Cπ -3

C= (3-3e^π)/π

F(x) = 3e^x -3sin(x) + (3-3e^π)/π x - 3

- Math -
**MathMate**, Friday, June 24, 2011 at 2:53pmf(0) = 0

f(π) = 0

Check if one of them is actually f'().

Assuming it's correct as above:

from:

f(x) =3e^x -3sin(x) + (3-3e^π)x/π - 3

We get:

f"(x)=3sin(x)+3e^x

f(0)=0

f(π)=0

which clearly satisfy all the given requirements. However, since there were two initial conditions for f(x), it is possible to have multiple solutions.

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