Mathematica Asked on June 20, 2021
"12.0.0 for Microsoft Windows (64-bit) (April 6, 2019)"
This can be solved by hand without too much difficulty. But
Solve[{s == 8/27 ((Sqrt[1 + (3/2 t)^2])^3 - 1), t > 0}, t, Reals]
results in
{{t -> ConditionalExpression[
Root[-16 s - 27 s^2 + 16 #1^2 + 36 #1^4 + 27 #1^6 &, 2], s > 0]}}
This works:
Module[{s},
s[t_] := 8/27 ((Sqrt[1 + (3/2 t)^2])^3 - 1);
InverseFunction[s]["t"]
]
Answered by Steven Thomas Hatton on June 20, 2021
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