How do you find the range of the function: #f(t) = 1 + 0.9 e ^(-0.02t)#?

1 Answer
Jul 24, 2018

Answer:

The range is #(1,+oo)#

Explanation:

The function is

#f(t)=1+0.9e^(-0.02t)#

The domain of this function is #t in RR#

That is

#t in (-oo, +oo)#

The limits are

#lim_(t->-oo)f(t)=lim_(t->-oo)(1+0.9e^(-0.02t))#

#=1+0.9e^(+oo)#

#=+oo#

#lim_(t->+oo)f(t)=lim_(t->+oo)(1+0.9e^(-0.02t))#

#=1+0.9e^(-oo)#

#=1+0#

#=1#

The range is #(1,+oo)#

graph{1+0.9e^(-0.02x) [-213.7, 213.6, -107, 106.9]}