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Consider the function [tex]\( f(t) = \sqrt{3t - 9} \)[/tex].

You cannot take the square root of a negative number, so [tex]\( 3t - 9 \)[/tex] must be zero or positive.

Determine the domain of [tex]\( f(t) \)[/tex].

Sagot :

GinnyAnsweridn
To find the conditions under which the function [tex]\( f(t) = \sqrt{3t - 9} \)[/tex] is defined, we need to ensure that the expression inside the square root is non-negative. Specifically, [tex]\( 3t - 9 \)[/tex] must be greater than or equal to zero because the square root of a negative number is not defined in the set of real numbers. Here’s the step-by-step process to find the restricted domain where the function is defined:

1. Expression Constraint:
[tex]\[ 3t - 9 \geq 0 \][/tex]

2. Solve for [tex]\( t \)[/tex]:
- Add 9 to both sides:
[tex]\[ 3t \geq 9 \][/tex]
- Divide both sides by 3:
[tex]\[ t \geq 3 \][/tex]

Thus, [tex]\( t \)[/tex] must be greater than or equal to 3 for the function [tex]\( f(t) = \sqrt{3t - 9} \)[/tex] to be defined.

However, an additional instruction is given in the problem stating that [tex]\( 3t - 9 \)[/tex] must be zero. Let’s solve for the specific value of [tex]\( t \)[/tex] where this condition holds true:

1. Set the Expression to Zero:
[tex]\[ 3t - 9 = 0 \][/tex]

2. Solve for [tex]\( t \)[/tex]:
- Add 9 to both sides:
[tex]\[ 3t = 9 \][/tex]
- Divide both sides by 3:
[tex]\[ t = 3 \][/tex]

Therefore, [tex]\( t = 3 \)[/tex] is the specific value where [tex]\( 3t - 9 \)[/tex] is zero.

### Summary
- The normal constraint to ensure the function [tex]\( \sqrt{3t - 9} \)[/tex] is defined is [tex]\( t \geq 3 \)[/tex].
- Given the additional requirement that [tex]\( 3t - 9 \)[/tex] must be zero, the specific solution is [tex]\( t = 3 \)[/tex].

So, the answer to the problem is:
[tex]\[ t = 3 \][/tex]
This is the value of [tex]\( t \)[/tex] where [tex]\( 3t - 9 \)[/tex] is zero.
Semsee45idn

Answer:

Domain interval notation:  [3, ∞)

Domain inequality notation:  t ≥ 3

Step-by-step explanation:

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

As we cannot take the square root of a negative number, the expression under the square root sign of function [tex]f(x)=\sqrt{3t-9}[/tex] must be non-negative, so zero or positive. Therefore:

[tex]3t-9\geq 0[/tex]

To determine the domain of f(t), solve the inequality:

[tex]3t-9\geq 0 \\\\\\3t-9+9\geq 0+9 \\\\\\3t\geq 9 \\\\\\\dfrac{3t}{3}\geq\dfrac{9}{3} \\\\\\t\geq 3[/tex]

Therefore, the domain of function f(t) is all real values of t greater than or equal to 3.

[tex]\Large\boxed{\boxed{\begin{array}{l}\underline{\textsf{Domain of $f(t)=\sqrt{3t-9}$}}\\\\\textsf{Interval notation:} \;\; \;\;\;[3, \infty) \\\textsf{Inequality notation:} \;\;t \geq 3 \end{array}}}[/tex]

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