Capacitors and Inductors at t=0
Intuitively , what you are seeing is the integration of an exponential waveform.

Solving for \$I_L(t),V_L(t)\$  , you know \$I_L(0)=0A,  V_L(0)=32V\$, what is \$I_L(t)\$ max ? ,

at t=0 V=32V all the step current is into the resistor then the inductor starts to integrate current as the resistor decays in current and voltage.


eventually the current slowly reaches a maximum dI/dt=o which means V=0 yet the current must be dissipated thru R so the voltage keep decaying to some negative lower voltage
as I decays to 0 the negative voltage also decays to 0

other thoughts.
In the time domain, for L,  V(t)= L dI(t)/dt  or you can integrate both sides for a current source as you have.
in the s domain for an ideal L,  Z(s)= s L
at  steady state s.
What happens when s=0 (dc !), You know a constant voltage starts with zero current when applied to an open circuit inductor and in theory ramps up to infinity, yet in practice the DCR series resistance limits current not shown.
So the steady impedance at DC is equivalent to 0 Ohms yet the transient impedance is infinite (meaning transient change in impedance starting up , or a change in voltage) because initially the current does not change.
This is simply the property of any integrator and the inductor integrates current with the applied voltage just like the capacitor integrates the opposite (charge voltage is integrated from current changes.) So the capacitor has the inverse impedance characteristics in the s domain to the inductor.
Now tackle your problem.
Given a Time domain current that steps from 0 to 8A then decays (-2t) exponentially to 0 with t/τ = 2t or τ=0.5 = &quot;tau&quot; the greek symbol used for exponential Time Constant
So what the L is the value ?

(sorry for puny pun)for T = 4 second exponential time constant.

Ask your Prof why show I(s) when the impedance is not constant in the s domain?
See what he /she says. (hehe) This is a time domain problem.
The s term in the L(s) may be confusing you because s domain is implies steady state spectrum (s) for which this is clearly not , it's a transient DC or &quot;baseband signal&quot; and the inductor impedance is constantly changing.

At t=0 the output voltage steps to 8A x 4Ω= 32V then exponential decays
As the inductor integrates up in current to a peak at which point dV/dt=0 and the thus the Voltage = 0 but the resistor has been draining the current, so the inductor current decays slower as required (4s) and the that negative dI/dt means the voltage swings below 0 (negative) but much less than the step voltage. - then - then voltage decays from -ve towards 0 as the inductor current decays exponentially to 0 with tau=L/R