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

1 Answer
Aug 28, 2015

Answer:

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]}