determine the domain of n(t) = sqrt(t/(1+t))

To determine the domain of the function \( n(t) = \sqrt{\frac{t}{1+t}} \), we need to consider the conditions under which the expression inside the square root is defined and non-negative.

1. The expression under the square root, \( \frac{t}{1+t} \), must be **non-negative** (i.e., greater than or equal to 0) because the square root is only defined for non-negative values.
   
2. Additionally, the denominator \( 1 + t \) cannot be zero, as division by zero is undefined.

### Step 1: Non-negativity Condition
For \( \frac{t}{1+t} \geq 0 \), the sign of the fraction depends on both the numerator \( t \) and the denominator \( 1+t \). Let's analyze this in two parts.

- **Numerator condition**: \( t \geq 0 \) (for the fraction to be non-negative).
- **Denominator condition**: \( 1 + t > 0 \) implies \( t > -1 \).

### Step 2: Combined Condition
From both conditions, we get:
- \( t \geq 0 \) from the numerator.
- \( t > -1 \) from the denominator.

The strictest condition comes from \( t \geq 0 \), so the domain of the function is \( t \geq 0 \).

### Final Answer:
The domain of \( n(t) = \sqrt{\frac{t}{1+t}} \) is:
\[
\boxed{[0, \infty)}
\]

Would you like a more detailed explanation or have any other questions?

Here are 5 related questions for further exploration:
1. What is the range of the function \( n(t) = \sqrt{\frac{t}{1+t}} \)?
2. How do you determine the domain of more complex functions with square roots and fractions?
3. What happens if we change the function to \( n(t) = \sqrt{\frac{1+t}{t}} \)?
4. How would you graph the function \( n(t) \)?
5. How can you apply these principles to functions with logarithms?

**Tip:** Always ensure both the square root and denominator are well-defined when determining the domain of functions involving these components.

Determine the domain of n(t) = sqrt(t/(1+t))

Learn how to find the domain of the function n(t) = sqrt(t/(1+t)). This problem involves understanding algebraic principles, square roots, and fraction inequalities, suitable for students in Grades 10-12. Step-by-step solutions provided.

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