f(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.
Can you help me with this? I did this many times and got different answers each time.
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
1 answer
1 answer