Q.If $V_{DS} = 4$ V and $V_{GS} = -1$ V, find the value of $\mu$.

The amplification factor $\mu$ of a JFET tells you how much more strongly the drain-source voltage can be varied than the gate-source voltage for the same change in drain current. It is defined as the ratio of a small change in drain-source voltage to the corresponding change in gate-source voltage, at constant drain current:

Being a ratio of two voltages, $\mu$ is a pure number with no units.

The three JFET parameters are connected. Starting from the definition and inserting the factor $\dfrac{\Delta I_D}{\Delta I_D}$:

$\mu = \frac{\Delta V_{DS}}{\Delta V_{GS}} \times \frac{\Delta I_D}{\Delta I_D} = \frac{\Delta V_{DS}}{\Delta I_D} \times \frac{\Delta I_D}{\Delta V_{GS}}$ …