Answer
$$
C \approx 0.2 \text { farads. }
$$
Work Step by Step
Use $$f^{\prime}(a)=\frac{1}{2}\left(\frac{f(a+h)-f(a)}{h}+\frac{f(a-h)-f(a)}{-h}\right)=\frac{f(a+h)-f(a-h)}{2 h}$$
Solving $i(4)=C v^{\prime}(4)+(1 / R) v(4)$ for $C$ yields
$$
C=\frac{i(4)-(1 / R) v(4)}{v^{\prime}(4)}=\frac{34.1-\frac{420}{200}}{v^{\prime}(4)}
$$
To compute $C,$ we first approximate $v^{\prime}(4) .$ Taking $h=0.1,$ we find
$$
v^{\prime}(4) \approx \frac{v(4.1)-v(3.9)}{0.2}=\frac{436.2-404.2}{0.2}=160
$$
Plugging this in to the equation above yields
$$
C \approx \frac{34.1-2.1}{160}=0.2 \text { farads. }
$$
