$y=\sin ^{-1}\left(\frac{1-x}{1+x}\right)$
Answer (1)
Find the domain of y = sin − 1 ( 1 + x 1 − x ) by ensuring − 1 ≤ 1 + x 1 − x ≤ 1 and x = − 1 , resulting in x ≥ 0 .
Calculate the derivative using the chain rule, obtaining d x d y = ( 1 + x ) x − 1 .
Analyze the derivative to determine that the function is decreasing on its domain.
Determine the range by considering the values of x in the domain [ 0 , ∞ ) , finding the range to be ( − 2 π , 2 π ] .
Explanation
Problem Analysis We are given the function y = sin − 1 ( 1 + x 1 − x ) . Our goal is to analyze this function, which includes finding its domain, derivative, and other relevant properties.
Domain of the Function First, let's determine the domain of the function. The arcsine function, sin − 1 ( u ) , is defined for − 1 ≤ u ≤ 1 . Therefore, we must have − 1 ≤ 1 + x 1 − x ≤ 1
Also, the denominator 1 + x cannot be zero, so x = − 1 . Let's analyze the two inequalities separately.
1 + x 1 − x ≤ 1 :
1 + x 1 − x − 1 ≤ 0 1 + x 1 − x − ( 1 + x ) ≤ 0 1 + x − 2 x ≤ 0 This inequality holds when -1"> x > − 1 or x ≥ 0 .
− 1 ≤ 1 + x 1 − x :
0 ≤ 1 + x 1 − x + 1 0 ≤ 1 + x 1 − x + ( 1 + x ) 0 ≤ 1 + x 2 This inequality holds when 0"> 1 + x > 0 , which means -1"> x > − 1 .
Combining these conditions, we find that the domain of the function is x ≥ 0 .
Derivative of the Function Next, let's find the derivative of the function using the chain rule. We have y = sin − 1 ( 1 + x 1 − x ) Let u = 1 + x 1 − x . Then y = sin − 1 ( u ) , and d u d y = 1 − u 2 1 . Also, d x d u = ( 1 + x ) 2 ( − 1 ) ( 1 + x ) − ( 1 − x ) ( 1 ) = ( 1 + x ) 2 − 1 − x − 1 + x = ( 1 + x ) 2 − 2 Therefore, d x d y = d u d y ⋅ d x d u = 1 − ( 1 + x 1 − x ) 2 1 ⋅ ( 1 + x ) 2 − 2 d x d y = ( 1 + x ) 2 1 − ( 1 + x ) 2 ( 1 − x ) 2 − 2 = ( 1 + x ) 2 ( 1 + x ) 2 ( 1 + x ) 2 − ( 1 − x ) 2 − 2 d x d y = ( 1 + x ) 2 1 + x ( 1 + x ) 2 − ( 1 − x ) 2 − 2 = ( 1 + x ) ( 1 + 2 x + x 2 ) − ( 1 − 2 x + x 2 ) − 2 d x d y = ( 1 + x ) 4 x − 2 = 2 ( 1 + x ) x − 2 = ( 1 + x ) x − 1 Thus, the derivative is d x d y = ( 1 + x ) x − 1 .
Analyzing the Derivative The derivative is d x d y = ( 1 + x ) x − 1 . Since x ≥ 0 , the derivative is always negative. This means that the function is decreasing on its domain.
Range of the Function To determine the range, we consider the values of x in the domain [ 0 , ∞ ) . When x = 0 , y = sin − 1 ( 1 + 0 1 − 0 ) = sin − 1 ( 1 ) = 2 π . As x approaches infinity, 1 + x 1 − x approaches − 1 , so y approaches sin − 1 ( − 1 ) = − 2 π . Therefore, the range of the function is ( − 2 π , 2 π ] .
Summary In summary, for the function y = sin − 1 ( 1 + x 1 − x ) :
Domain: x ≥ 0
Derivative: d x d y = ( 1 + x ) x − 1
The function is decreasing on its domain.
Range: ( − 2 π , 2 π ]
Examples
The arcsine function and its transformations are used in various fields such as physics and engineering. For example, in signal processing, the arcsine function can be used to model phase demodulation. Understanding the domain, range, and derivative of such functions is crucial for analyzing and designing systems that involve these functions. Additionally, in computer graphics, arcsine functions are used in lighting models and texture mapping to create realistic visual effects.
