What is the domain of $y = \frac{3}{2} arcsin (2 x + .5)$ $y=3/2 arcsin(2x+.5)$ ?

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Answer 1 · 2018-01-17T09:00:57

$- \frac{3}{4} \leq x \leq \frac{1}{4}$ $-3/4 <= x <= 1/4$

The domain of ${sin}^{- 1} (u)$ $sin^-1(u)$ is $- 1 \leq u \leq 1$ $-1<=u<=1$ , so for this problem we need to solve:

$- 1 \leq 2 x + \frac{1}{2} \leq 1$ $-1<=2x+1/2<=1$

subtracting $\frac{1}{2}$ $1/2$ from all parts:

$- \frac{3}{2} \leq 2 x \leq \frac{1}{2}$ $-3/2 <= 2x <= 1/2$

dividing through by 2:

$- \frac{3}{4} \leq x \leq \frac{1}{4}$ $-3/4 <= x <= 1/4$

What is the domain of y=3/2 arcsin(2x+.5)y=32arcsin(2x+.5)y=3/2 arcsin(2x+.5)?