The function `f(x)=sin^(-1)(cosx)` is

Options

A. discontinuous at x = 0
B. continuous at x = 0
C. differentiable at x = 0
D. None of these

Answer

To determine the properties of the function \( f(x) = \sin^{-1}(\cos x) \), we will analyze its continuity and differentiability step by step.

### Step 1: Understanding the Function
The function \( f(x) = \sin^{-1}(\cos x) \) involves the inverse sine function applied to the cosine of \( x \). The cosine function oscillates between -1 and 1, which is within the range of the inverse sine function.

### Step 2: Finding the Domain
The domain of \( f(x) \) is determined by the values of \( \cos x \) since \( \sin^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Since \( \cos x \) always lies in this interval, the domain of \( f(x) \) is all real numbers.

### Step 3: Analyzing Continuity
To check for continuity at a point, we need to verify if:
\[ \lim_{x \to c} f(x) = f(c) \]
for some point \( c \). Let's check continuity at \( x = 0 \):
\[ f(0) = \sin^{-1}(\cos(0)) = \sin^{-1}(1) = \frac{\pi}{2} \]
Now, we compute the limit as \( x \) approaches 0:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \sin^{-1}(\cos x) = \sin^{-1}(\cos(0)) = \sin^{-1}(1) = \frac{\pi}{2} \]
Since \( \lim_{x \to 0} f(x) = f(0) \), the function is continuous at \( x = 0 \).

### Step 4: Analyzing Differentiability
To check differentiability at \( x = 0 \), we need to find the derivative \( f'(x) \) and check if it exists at \( x = 0 \). We will use the chain rule:
\[ f'(x) = \frac{d}{dx} \sin^{-1}(\cos x) = \frac{1}{\sqrt{1 - (\cos x)^2}} \cdot (-\sin x) = \frac{-\sin x}{\sqrt{1 - \cos^2 x}} = \frac{-\sin x}{|\sin x|} = -1 \text{ or } 1 \]
The sign of \( \sin x \) changes depending on the interval. Therefore, the derivative does not have a unique value at \( x = 0 \) (it approaches -1 from the left and 1 from the right). Hence, \( f'(0) \) does not exist.

### Conclusion
- The function \( f(x) = \sin^{-1}(\cos x) \) is continuous at \( x = 0 \).
- The function \( f(x) \) is not differentiable at \( x = 0 \).

### Final Answer
The function \( f(x) = \sin^{-1}(\cos x) \) is continuous but not differentiable at \( x = 0 \).