Involves exponential and log functions

Question

The temperature, T ( Degree C), of a quenched steel plate at time t is given by
T = f(t) = 10 + 85/1 + e^ (t−60)/10
where 0 ≤ t ≤ 100 measured in seconds.
(a) We want to know when the temperature will reach a certain value. Solve for the time t in terms of the temperature T, and hence find the formula for the inverse function f −1 (t).
(b) Write down the domain of f −1 (t).
(c) Plot f(t) from t = 0 to t = 100 and then plot f −1 (t) on the same axes but only for the values of t in the domain of the inverse function found in (b).
(d) Describe the geometric relationship that you can see when you consider both f(t) and f −1 (t).

Answer 1

I assume your function is f(t) = 10 + 85 / (1 + e^(t-60)/10).

 

 

(a)

 

We'd like to find the inverse of the given function by solving for t.

 

T - 10 = 85 / (1 + e^(t-60)/10)

(T - 10)(1 + e^(t-60)/10) = 85

1 + e^(t-60)/10 = 85/(T - 10)

e^(t-60)/10 = -1 + 85/(T - 10)

ln(-1 + 85/(T-10)) = (t-60)/10

10ln(-1 + 85/(T-10)) = t - 60

10ln(-1 + 85/(T-10)) + 60 = t

 

So, f^-1(t) = 10ln(-1 + 85/(t-10)) + 60

 

 

For (b) and (c), you'll need to plot the two functions.