Question

Having trouble solving this practice problems from my ACT prep guide

Answer 1

Notice that sin(x) is not a one-to-one function since

`\sin x=\sin (x+2\pi)`

(The period of the sine function is 2pi)

On the other hand, a function needs to be one-to-one in order to be invertible.

Thus, the answer to the first gap is 'is not one-to-one'.

As for the second gap, notice that

`\begin{gathered} \sin (0)=\sin (\pi)=0 \\ \sin (-(\pi)/(2))=-1,\sin ((\pi)/(2))=1 \end{gathered}`

The maximum interval at which f(x) is one to one is

`\text{one}-to-\text{one}=\mleft\lbrace x\in\R|-(\pi)/(2)\le x\le(\pi)/(2)\mright\rbrace=\lbrack-(\pi)/(2),(\pi)/(2)\rbrack`

This is the maximum interval that verifies the horizontal-line test.

The answer to the second gap is [-pi/2,pi/2]

Notice that if the domain of the inverse function is [-1,1]; then the domain of such function is [-pi/2,pi/2]