The function f(x) is defined on the interval (0,1). What are the domains of definition of the following functions.
(a) `f(3x^(2)), (b) f(x-5) , (c) f(tan x)?`

Answer

To find the domains of the functions \( f(3x^2) \), \( f(x-5) \), and \( f(\tan x) \), we need to ensure that the arguments of \( f \) fall within the interval (0, 1), as that is the defined domain for \( f(x) \).

### Solution Steps:

1. **Finding the Domain of \( f(3x^2) \)**:
- We need \( 3x^2 \) to be in the interval (0, 1).
- This gives us the inequality:
\[ 0 < 3x^2 < 1 \]
- Dividing the entire inequality by 3:
\[ 0 < x^2 < \frac{1}{3} \]
- Taking the square root (considering only positive values since \( x \) is in (0, 1)):
\[ 0 < x < \frac{1}{\sqrt{3}} \]
- Therefore, the domain of \( f(3x^2) \) is:
\[ (0, \frac{1}{\sqrt{3}}) \]

2. **Finding the Domain of \( f(x-5) \)**:
- We need \( x-5 \) to be in the interval (0, 1).
- This gives us the inequality:
\[ 0 < x - 5 < 1 \]
- Solving the inequalities:
- From \( 0 < x - 5 \):
\[ x > 5 \]
- From \( x - 5 < 1 \):
\[ x < 6 \]
- Therefore, the domain of \( f(x-5) \) is:
\[ (5, 6) \]

3. **Finding the Domain of \( f(\tan x) \)**:
- We need \( \tan x \) to be in the interval (0, 1).
- The function \( \tan x \) is positive in the first quadrant (0 to \( \frac{\pi}{2} \)) and has a range of (0, 1) in that interval.
- We can find the values of \( x \) for which \( \tan x < 1 \):
- This occurs when:
\[ x < \frac{\pi}{4} \]
- Since \( \tan x \) is periodic with a period of \( \pi \), we can express the domain as:
\[ x \in \left( n\pi, n\pi + \frac{\pi}{4} \right) \quad \text{for integers } n \]
- Therefore, the domain of \( f(\tan x) \) is:
\[ \bigcup_{n \in \mathbb{Z}} \left( n\pi, n\pi + \frac{\pi}{4} \right) \]

### Final Results:
- (a) Domain of \( f(3x^2) \): \( (0, \frac{1}{\sqrt{3}}) \)
- (b) Domain of \( f(x-5) \): \( (5, 6) \)
- (c) Domain of \( f(\tan x) \): \( \bigcup_{n \in \mathbb{Z}} \left( n\pi, n\pi + \frac{\pi}{4} \right) \)