If $f(x)=f^{\prime}(x)$ and $f(1)=2$, then $f(3)$ is

Given the differential equation $f(x) = f'(x)$, we can write it as $\dfrac{df}{dx} = f(x)$.

Separating the variables, we have $\dfrac{df}{f} = dx$.

Integrating both sides, we get $\ln|f(x)| = x + C$, which implies $f(x) = Ae^x$ for some constant $A$.

Using the initial condition $f(1) = 2$, we substitute $x=1$: $2 = Ae^1$, which gives $A = \dfrac{2}{e}$.

Thus, the function is $f(x) = \left(\dfrac{2}{e}\right)e^x = 2e^{x-1}$.

To find $f(3)$, we substitute $x=3$: $f(3) = 2e^{3-1} = 2e^2$.

Answer: $2 e^2$