This function is housed by a square root. So the function is only defined when the resulting output is defined.

yea and since the variable t is in the squareroot, only the entire value of the squareroot can be manipulated. the lowest value the squareroot can go is 0, then f(t) will be 5 in this case. so this is ur max.
as t increases, the squareroot value increases, and thus f(t) decreases.
so your range should be f(t) <= 5

sorry if it’s confusing just my way of getting the range. hope it helps

f(t) = 5 - 3 sqrt (t + 4)

Square roots are notoriously known for causing trouble to negative numbers. The square of any real number cannot be negative (can only be zero or positive); conversely, a negative number will not produce a real number under the influence of a square root.

So, for f(t) to exist,
t + 4 >= 0 (I am using phone to type instead of computer so I have no access to the greater than or equal to sign)
t >= -4

We can put the domain (input) as [-4, infinity) (have you learned to express notation as such)?

When t = -4, the resulting output is 5.
For all values of t, the square root has a resulting positive output, so the overall resulting output is lower (because 5 minus a positive number is lower than 5).

We can put the range (output) as (-infinity, 5].

f(t) = 5 - 3√(t+4)

How to reason it out :

① Real value of a square root is only obtained when the term inside ≥ 0 .

A real value of a square root is always ≥ 0 (unless you put a negative sign outside)

Eg. √4 = 2, √0 = 0, √10000 = 100


Otherwise, you will get imaginary numbers, eg. √-1 = i


So, t + 4 ≥ 0

And so , t ≥ -4

t just needs to be bigger than or equal to -4.


Either express the domain in interval notation : [-4,∞)

(Square brackets mean the value of the lower bound is included, curved brackets mean that it doesn't. In this case for infinity you'll put curved as it is never attainable)

or

set of values (notation) like you did : {t ∈ R : t ≥ 4}


② f(t) = 5 - 3√(t+4)

As t increases from -4 onwards (we can say it tends to positive infinity) , so will t + 4.

The square root √(t + 4) gets larger and so does 3√(t + 4)

5 - 3√(t+4) gets smaller and smaller, and never really ends (we can say it tends to negative Infinity)

smallest possible value of the √ is 0 so largest possible value of f(t) is 5

It can be represented by a few inequalities :

t + 4 ≥ 0
√(t+4) ≥ 0
3√(t + 4) ≥ 0
-3√(t + 4) ≤ 0
5 - 3√(t + 4) ≤ 5

f(t) ≤ 5


So the set of values for the range is :
{f(t) ∈ R: f(t) ≤ 5｝

Interval notation will be : (-∞, 5]