Problem 85 - Algebra and Trigonometry

Question

Determine the domain and the range of each function. $$ f(x)=\sin \left(\sin ^{-1} x\right) $$

Step-by-Step Solution

Answer

The domain of the function \( f(x)=\sin \left(\sin ^{-1} x\right) \) is \([-1,1]\) and the range is also \([-1,1]\).

Step 1: Define given function

The given function is \( f(x)=\sin \left(\sin ^{-1} x\right) \). This function is a composition of the sine function and its inverse.

Step 2: Find the Domain

The domain of a function is the set of all possible input values for which the function is defined. Since the inside function is \(\sin ^{-1} x\), we can say that the domain is the values for which \(\sin ^{-1} x\) is defined. We know that the \( \sin ^{-1} x \) is defined for all \( x \) in the interval \([-1,1]\). Therefore, the domain of the given function \( f(x)=\sin \left(\sin ^{-1} x\right) \) is \([-1,1]\).

Step 3: Find the Range

The range of a function is the set of all possible output values. Here, the output is determined by the outside function \( \sin(x) \). We know that \( \sin(x) \) can take any value between -1 and 1, for all \( x \). Therefore, the range of \( f(x)=\sin \left(\sin ^{-1} x\right) \) is also \([-1,1]\).

Key Concepts

Inverse Functions

Inverse functions are an important mathematical concept that essentially reverse the effect of the original function. If you have a function, say, \( f(x) \), its inverse function, denoted as \( f^{-1}(x) \), will map the output back to the original input. This means if \( f(a) = b \), then \( f^{-1}(b) = a \). However, not all functions have inverses. For a function to have an inverse, it must be bijective, i.e., both injective and surjective.

In the context of the exercise, we are dealing with the sine function and its inverse. The inverse of the sine function is called arcsine, written as \( \sin^{-1}(x) \). This particular inverse is only defined on the interval \([-1, 1]\), meaning it can only accept input values within this range. Thus, understanding inverse functions helps us identify the valid domain of \( \sin^{-1}(x) \), which in this case, is also the domain for the composition involving \( f(x)=\sin(\sin^{-1}(x)) \).

Sine Function

The sine function is a fundamental element of trigonometry and appears frequently in various domains of mathematics and engineering. It is periodic, meaning it repeats its values in regular intervals of \( 2\pi \). The function \( \sin(x) \) maps any real number \( x \) to a value between -1 and 1.

For the sine function, its range is always

• -1 as its minimum value
• 1 as its maximum value

This is crucial when considering the range of any function composed of a sine function, like the one in our exercise. Even though the input range is restricted by the domain of the inverse function, \( \sin^{-1}(x) \), the output doesn’t need to be, because the values \( \sin(x) \) can independently recapture all values in the interval \([-1,1]\). So, in this problem, knowing the behavior and range of the sine function helps us easily deduce the range of \( f(x) = \sin(\sin^{-1}(x)) \).

Composition of Functions

The composition of functions is a higher-level concept in mathematics where two functions are combined to form a new function. If you have two functions, \( g(x) \) and \( h(x) \), the composition \( f(x) = g(h(x)) \) implies that you first apply \( h \) and then apply \( g \) to the result.

In this exercise, the composition is illustrated as \( f(x)=\sin(\sin^{-1}(x)) \). This means we are applying the \( \sin^{-1}(x) \) (arcsin) function first, which is only valid for inputs in \([-1,1]\), thus initially setting the domain. Following that, the sine function is applied to the arcsine results. This simplifies beautifully because \( \sin(\sin^{-1}(x)) = x \) for all \( x \) in the domain of arcsine,

• creating an identity for the function

Hence, the understanding of function composition helps unravel the apparent complexity of such functions by looking at their structure in layers: an inner function followed by an outer function.