Answer :

grujan50

Answer:

Correct answer:  sin x ⇒D(x) :  [- π/2, π/2] ; sin⁻¹x ⇒ CD(x) :  [- π/2, π/2]

Step-by-step explanation:

In order for the function sin x to have an inverse function sin⁻¹x due to the monotony, the domain is taken D(x) :  [- π/2, π/2] and the range of sin⁻¹x  is CD(x) :  [- π/2, π/2].

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Using inverse function concepts, it is found the squares are completed with: [tex][0,2\pi][/tex]

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  • The inverse function only exists if: [tex]f(a) = f(b) \rightarrow a = b[/tex], that is, for each output, there is only one input.
  • The sine function, as given by the graph at the end of this question, is periodic with period [tex]2\pi[/tex], in the interval [tex][0,2\pi][/tex], which means that after that, the values repeat. Thus, for the inverse function to exist, only one period, that is [tex][0,2\pi][/tex], is considered.
  • Since the output of the inverse function is the input of the original function, all functional values of [tex]f(x) = \sin^{-1}{x}[/tex] are on the interval [tex][0,2\pi][/tex]

A similar question is given at https://brainly.com/question/23339681

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