What is the range of this f(x) = sin^-1(2x - 1)?

Question 7

EasyOriginal question

Let $f(x) = \sin^{-1}(2x - 1)$ .

What is the range of this $f(x) = \sin^{-1}(2x - 1)$ ?

\left[0, \pi\right]

Correct answerYour answer

\left[0, \frac{\pi}{2}\right]

Correct answerYour answer

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Correct answerYour answer

\left[-\pi, \pi\right]

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

As $x$ runs over $[0, 1]$ , $2x - 1$ runs over all of $[-1, 1]$ .
So $f$ takes every value of $\sin^{-1}$ : $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

Worth noting

The inner $2x - 1$ changes the domain, not the range, because it covers $[-1, 1]$ exactly.

Ten questions from Inverse Trigonometric Functions, at your own pace.

Ten questions from Inverse Trigonometric Functions, at your own pace.