I am after the formula for the inverse function f-1(t)

Question

The original function is f(t) = V = 2+48/(1+e^{-(t-20)³/1000})

I have plotted this original and for 0>=t>=40 and get a upward snaking line. For every inverse I derive I get a completely different plot and is not a reflection on the line y=t. Can someone please advise if my original plot is incorrect or derive the correct inverse of this function? 

Answer 1

V = f(t) = 2 + 48 / (1+e^{-(t-20)^3/1000)}

 

Switch V and t:  t = 2 + 48 / (1+e^{-(V-20)^3/1000})

 

Solve for V:  t-2 = 48/(1+e^{-(V-20)^3/1000})

 

                   1+e^{-(V-20)^3/1000} = 48/(t-2)

 

                       e^{-(V-20)^3/1000} = 48/(t-2) - 1

 

                      -(V-20)^3/1000 = ln[48/(t-2) - 1]

 

                      -(V-20)³ = 1000ln[48/(t-2) - 1]

 

                        (V-20)³ = -1000ln[48/(t-2) - 1]

 

                         V-20 = -10{ln[48/(t-2) - 1]}^1/3

 

                            V = f^-1(t) = 20 - 10{ln[48/(t-2) - 1]}^1/3