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Question

sketch a graph of the function and compare the graph of g with the graph of f(x) = arcsin x. g(x) = arcsin (x - 1)

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Answered 2 years ago
Answered 2 years ago
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Given funcitons

f(x)=arcsinxf\left(x\right)=\arcsin x

g(x)=arcsin(x1)g\left(x\right)=\arcsin \left(x-1\right)

We need to graph this functions and to compare the graph of gg with the graph of ff.

We know that

The domain of f is : 1x1\color{#c34632}{\text{The domain of $f$ is : } -1 \leq x \leq 1}

The range of f is : π2yπ2\color{#c34632}{\text{The range of $f$ is : } -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}}

Therefore,

The domain of g is : 2x0\color{#4257b2}{\text{The domain of $g$ is : } -2 \leq x \leq 0}

The range of g is : π2yπ2\color{#4257b2}{\text{The range of $g$ is : } -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}}

For x=2x=2 :

g(2)=arcsin(21)=arcsin1=π2g\left( 2 \right)=\arcsin \left(2-1\right)=\arcsin 1=\frac{\pi}{2}

For x=0x=0 :

f(0)=arcsin0=0f\left(0\right)=\arcsin 0=0

g(0)=arcsin(01)=arcsin(1)=π2g\left(0\right)=\arcsin \left(0-1\right)=\arcsin \left(-1\right)=-\frac{\pi}{2}

For x=±1x=\pm 1 :

f(±1)=arcsin(±1)=±π2f\left(\pm 1\right)=\arcsin \left(\pm 1 \right)=\pm \frac{\pi}{2}

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