How do you find the domain and range of $arcsin (3 x + 1)$ $arcsin(3x+1)$ ?

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How do you find the domain and range of $arcsin (3 x + 1)$ $arcsin(3x+1)$ ?

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Answer 1 · 2017-10-29T10:12:32

Domain: $x \in [- \frac{2}{3}, 0]$ $x in [-2/3,0]$
Range: $y \in [- \frac{π}{2}, \frac{π}{2}]$ $y in [-pi/2, pi/2]$

Explanation:

It is easy if you treat it as a transformation of your basic $arcsin$ $arcsin$ graph, knowing that the domain is $[- .1, 1]$ $[-.1,1]$ and the range is $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ .

Then taking your function $y = a f (b x + c) + d$ $y = a f(bx+c) + d$ , transpose it so that the actual function is on its own: $\frac{y - d}{a} = f (b x + c)$ $(y-d)/a = f(bx+c)$ and substitute each $y$ $y$ and $x$ $x$ transformation into the range/domain, and solve as an inequality.

Hence you get $- \frac{π}{2} \leq y \leq \frac{π}{2}$ $-pi/2<=y<=pi/2$ and $- 1 \leq 3 x + 1 \leq 1$ $-1<=3x+1<=1$ .

Of course the first one needs no solving; we already have the range.

Rearranging the second inequality we get the domain:
$- \frac{2}{3} \leq x \leq 0$ $-2/3<=x<=0$

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How do you find the domain and range of $arcsin (3 x + 1)$ $arcsin(3x+1)$ ?

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How do you find the domain and range of arcsin(3x+1)arcsin(3x+1)?

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How do you find the domain and range of $arcsin (3 x + 1)$ $arcsin(3x+1)$ ?