The function `y=sin^(-1)` (1+x) increases in `-2 lt x lt 0`. Is it true?

Answer

To determine whether the function \( y = \sin^{-1}(1 + x) \) is increasing in the interval \( (-2, 0) \), we will follow these steps:

### Step 1: Find the derivative of the function
To check if the function is increasing, we need to find the derivative of \( y \) with respect to \( x \). The derivative of \( y = \sin^{-1}(u) \) where \( u = 1 + x \) is given by:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
Here, \( \frac{du}{dx} = 1 \) since \( u = 1 + x \).

Substituting \( u \) into the derivative:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (1 + x)^2}} \cdot 1 = \frac{1}{\sqrt{1 - (1 + 2x + x^2)}} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{-x^2 - 2x}} \]

### Step 2: Determine when the derivative is non-negative
The function \( y \) is increasing when \( \frac{dy}{dx} \geq 0 \). This occurs when the expression inside the square root is positive:
\[ -x^2 - 2x \geq 0 \]
Rearranging gives:
\[ x^2 + 2x \leq 0 \]

### Step 3: Factor the inequality
Factoring the left-hand side:
\[ x(x + 2) \leq 0 \]

### Step 4: Find the critical points
The critical points occur when:
\[ x = 0 \quad \text{or} \quad x + 2 = 0 \Rightarrow x = -2 \]

### Step 5: Test intervals around the critical points
We will test the intervals determined by the critical points \( -2 \) and \( 0 \):
1. For \( x < -2 \) (e.g., \( x = -3 \)):
\[ (-3)(-3 + 2) = (-3)(-1) = 3 \quad (\text{positive}) \]
2. For \( -2 < x < 0 \) (e.g., \( x = -1 \)):
\[ (-1)(-1 + 2) = (-1)(1) = -1 \quad (\text{negative}) \]
3. For \( x > 0 \) (e.g., \( x = 1 \)):
\[ (1)(1 + 2) = (1)(3) = 3 \quad (\text{positive}) \]

### Step 6: Conclusion
The function \( y = \sin^{-1}(1 + x) \) is increasing in the interval \( (-2, 0) \) because the derivative \( \frac{dy}{dx} \) is non-negative in that interval.

Thus, the statement that the function increases in the interval \( (-2, 0) \) is **true**.