Is L(t)=1-e^-t an exponential function?

1 Answer
Bill K.
Aug 28, 2015

Yes and no. It's not a "pure" exponential function of the form f(t)=ae^{kt}, but it's an transformed version of an exponential function.

Explanation:

Exponential functions are those of the form f(t)=ae^{kt}=a*b^{t}, where b=e^{k} (and k=ln(b)).

The function f(t)=e^{-t} is therefore an exponential (decay) function with a=1 and k=-1.

The function g(t)=-f(t)=-e^{-t} is also an exponential function with a=-1 and k=-1, and its graph is transformed from the graph of f by reflection across the horizontal axis.

The function h(t)=1+g(t)=1-f(t)=1-e^{-t} takes the graph of g and shifts it up by 1 unit.

Here's what the graph of h looks like. It's increasing, concave down, and has a horizontal asymptote y=1 as t->+infty.

graph{1-e^(-x) [-10, 10, -5, 5]}

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