[SOLVED] Find a function of the form...?

**fardeen_gen** · May 3rd 2009, 08:57 PM

Find a function of the $f(x) = \lambda e^{2x} + \mu e^{x} - \nu x$ where $\lambda, \mu, \nu$ are independent of $x$ and $f(0) = -1$ , $f'(\log_{e} 2) = 30$ and $\int_{0}^{\log_{e} 4} \{f(x) + \nu x\} dx = 28.5$

**sirsosay@gmail.com** · May 3rd 2009, 10:00 PM

Just need to make a system of equations and substitute.

lamda=L mew=M v=V

f(0)=Le^(2*0)+M*e^(0)-v*0=-1
L+M=-1

f'(x)= 2*L*e^(2x)+M*e^(x) -V
f'(ln(2))=8*L+2*M-V=30

integral(L*e^(2*x)+M*e^(x),x,0,ln(4))

solve(L/2*e^(2*X)+M*e^x),x,0,ln(4))
8L+4M-(L/2+M)=28.5

System of equations:
1*L+1*M+0*V=-1
8*L+2*M-1*V=30
15/2*L+3*M=28.5

**fardeen_gen** · May 3rd 2009, 10:03 PM

Answer:

Spoiler:

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