Let f(t)=1+.9e^-t with t ≥ 0.?
(a) find
lim
----------- f(t).
1→ ∞
(b) show that the function is decreasing for all t.
(c) what is the range of the function? justify your answer.
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(a) find
lim
----------- f(t).
1→ ∞
(b) show that the function is decreasing for all t.
(c) what is the range of the function? justify your answer.
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Verified answer
a) f(t) = 1 + .9e^-t
f(t) = 1 + .9/e^t
As t → ∞, you get 1 + 0 = 1.
b) You need to show that the first derivative is always negative.
f'(t) = -.9e^-t
f'(t) = -.9/e^t
f'(t) never equals zero.
e^t is always positive.
-.9 is always negative.
Therefore f'(t) is always negative.
c) The range is (1,∞).
EDIT: I missed the "t ≥ 0" part.
f(0) = 1 + .9e^0 = 1 + .9 = 1.9. This is the largest that f(t) could be.
The range is (1, 1.9].