The function f(x) is defined on the interval [0,1]. What are the domains of definition of functions.
(a) `f(sin x), (b) f(2x+3)?`

Answer

To find the domains of the functions \( f(\sin x) \) and \( f(2x + 3) \) given that \( f(x) \) is defined on the interval \([0, 1]\), we will analyze each function step by step.

### Step 1: Determine the domain of \( f(\sin x) \)

1. **Identify the range of \( \sin x \)**:
- The sine function, \( \sin x \), oscillates between -1 and 1 for all \( x \).
- Therefore, the range of \( \sin x \) is \([-1, 1]\).

2. **Restrict to the interval where \( f(x) \) is defined**:
- Since \( f(x) \) is defined on \([0, 1]\), we need to find where \( \sin x \) falls within this interval.
- We need \( \sin x \) to be in the interval \([0, 1]\).

3. **Find the values of \( x \) for which \( \sin x \) is in \([0, 1]\)**:
- The sine function is non-negative (i.e., \( \sin x \geq 0 \)) in the intervals:
- \( [0, \pi] \) for \( x \) in radians.
- Therefore, the domain of \( f(\sin x) \) is \( x \in [0, \pi] \).

### Step 2: Determine the domain of \( f(2x + 3) \)

1. **Set up the inequality for \( f(2x + 3) \)**:
- We need \( 2x + 3 \) to be in the interval \([0, 1]\).
- This gives us the inequality:
\[ 0 \leq 2x + 3 \leq 1 \]

2. **Solve the inequalities**:
- Start with the left side:
\[ 2x + 3 \geq 0 \implies 2x \geq -3 \implies x \geq -\frac{3}{2} \]
- Now solve the right side:
\[ 2x + 3 \leq 1 \implies 2x \leq -2 \implies x \leq -1 \]

3. **Combine the results**:
- The combined inequalities give us:
\[ -\frac{3}{2} \leq x \leq -1 \]
- Therefore, the domain of \( f(2x + 3) \) is \( x \in [-\frac{3}{2}, -1] \).

### Final Answer:
- The domain of \( f(\sin x) \) is \( [0, \pi] \).
- The domain of \( f(2x + 3) \) is \( [-\frac{3}{2}, -1] \).