too bad you didn't bother to show your work.
f" = 4e^x - 9sinx
f' = 4e^x + 9cosx + c1
f = 4e^x + 9sinx + c1*x + c2
using the two values of f(x), we have
4+0+0+c2 = 3 ==> c2 = -1
4e^(π/2) + 9 + c1 * π/2 -1 = 0
c1 = -(8+4e^(π/2))/(π/2) = -8/π (2+e^(π/2))
f(x) = 4e^x + 9sinx - 8/π (2+e^(π/2)) x - 1
f''(x)= 4e^x-9sin(x) f(0)=3 f(pi/2)=0
I am to find f(x); but I am confused about how to do that. Could I please get some help? I think I have to find f'(x) and then go backwards from there, but my answer doesn't seem to work.
2 answers
Ah, thank you! My apologies for not showing my work; I got f(x)= 4e^x+9sinx+cx but forgot that it would be cx+c2 so I was just confused on how to solve it with the two f(x) values given. Thanks again!