Find domain and range

Question

Answer 1

Oops!! After looking at the answer provided by Mark M., I realize I wrote down the wrong function and thereby answered a different question. I'll leave my answer here in the hope that it may have some instructive value, but if you want the domain and range of g(t) = √(25 + 4^–t) you should look at Mark M.'s answer instead of mine.

Domain and range of g(t) = √(25 – 4^–t)

Domain

The domain will include all values of t where 25 – 4^–t ≥ 0 which means

25 ≥ 4^–t which can be rewritten 25 ≥ e^–tln4 (since 4^–t = (e^ln4)^–t = e^–tln4)

Taking the natural logarithm of both sides gives

ln25 ≥ –tln4 which means that ln25/ln4 ≥ –t which simplifies to

ln5/ln2 ≥ –t which can be rewritten as t ≥ –ln5/ln2

–ln5/ln2 ≤ t < ∞ is our domain, which in interval notation is

[–ln5/ln2,∞)

Range

As t → ∞, 4^–t → 0 and g(t) → √(25) = 5 (horizontal asymptote at 5)

At t = –ln5/ln2, 4^–t = 25 and g(t) = √(25 – 25) = 0 and this is the minimum value of g(t)

0 ≤ g(t) < 5 for all values of t

So the range of g(t) is [0,5)

Answer 2

25 + 4^-t = 25 + (1/4)^t

(1/4)^t > 0 for all t. So, 25 + (1/4)^t > 25 for all t.

f(t) = √ (25 + (1/4)^t)

Domain: (-∞, ∞) Range: (5, ∞)