Đáp án: $f(0)=e^{\dfrac{-5}{9}}$
Giải thích các bước giải:
$f'(x)=f(x).\sqrt{3x+1}$
$\rightarrow \dfrac{f'(x)}{f(x)}=\sqrt{3x+1}$
$\rightarrow \int\dfrac{f'(x)}{f(x)}dx=\int \sqrt{3x+1}dx$
$\rightarrow lnf(x)=\dfrac{2}{9}.(3x+1)^{\dfrac{3}{2}}+C$
Do $f(1)=e\rightarrow lnf(1)=\dfrac{2}{9}.(3.1+1)^{\dfrac{3}{2}}+C$
$\rightarrow C=-\dfrac{7}{9}$
$\rightarrow lnf(x)=\dfrac{2}{9}.(3x+1)^{\dfrac{3}{2}}-\dfrac{7}{9}$
$\rightarrow lnf(0)=\dfrac{-5}{9}\rightarrow f(0)=e^{\dfrac{-5}{9}}$