f(t)=3sqrt(2t-1)
why domain is all real #'s ?
3sqrt(2t-1) >= 0
( 3sqrt(2t-1) )^3 >= (0)^3
t>= 1/2
?
To understand why the domain of the function f(t) = 3√(2t - 1) is all real numbers, let's analyze the inequality you wrote down.
You correctly rewrote the inequality from 3√(2t - 1) ≥ 0. To solve this inequality, you raised both sides to the power of 3 (cubed both sides).
(3√(2t - 1))^3 ≥ (0)^3 simplifies to:
27(2t - 1) ≥ 0,
which further simplifies to:
54t - 27 ≥ 0,
or:
54t ≥ 27.
Now, we can divide both sides of the inequality by 54 to isolate t:
t ≥ 27/54,
which simplifies to:
t ≥ 1/2.
Therefore, we have determined that the inequality is satisfied when t is greater than or equal to 1/2.
To find the domain of the function f(t) = 3√(2t - 1), we need to consider the value inside the square root, 2t - 1.
For the square root to be defined, its argument (2t - 1) must be greater than or equal to 0 (since the square root of a negative number is not real).
Since the argument 2t - 1 is greater than or equal to 0 when t ≥ 1/2 (as we found earlier), the domain of the function includes all real numbers. Therefore, the domain of f(t) = 3√(2t - 1) is all real numbers.
if f(t) = 3√(2t-1) the domain is not all real numbers. You need
2t-1 >= 0
t >= 1/2
so the domain is [1/2,∞)
You are correct