Answer
\[
f(x)=\frac{7}{9}+\frac{2}{9}(1+3 x)^{\frac{3}{2}}
\]
Work Step by Step
We have the slope of the tangent line at a point $(x, y)$ on the curve $y=f(x)$ is: $\sqrt{1+3 x}$
This means
\[
\left.\frac{d f}{d x}\right|_{(x, y)}=\sqrt{1+3 x}
\]
We have to find the $f$
So, $f=\int \sqrt{1+3 x} d x=\int(1+3 x)^{\frac{1}{2}} d x$
By using the subtituation $u=1+3 x \Rightarrow d u=3 d x \Rightarrow \frac{d u}{3}=d x$
\[
\begin{aligned}
\int(1+3 x)^{\frac{1}{2}} d x &=\frac{1}{3} \int u^{\frac{1}{2}} d u \\
&=\frac{2}{9} u^{\frac{3}{2}}+C \\
&=\frac{2}{9}(1+3 x)^{\frac{3}{2}}+C
\end{aligned}
\]
The curve passes through the point (0,1)
$\therefore f(0)=1 \Rightarrow 1=\frac{2}{9}(1+3(0))^{\frac{3}{2}}+C$
$\Rightarrow 1=\frac{2}{9}+C \Rightarrow C=1-\frac{2}{9}=\sqrt{\frac{7}{9}}$
\[
\ f(x)=\frac{7}{9}+\frac{2}{9}(1+3 x)^{\frac{3}{2}}
\]
