What is the range of f(x)=arcsin(3x+1) ?

The range of f(x)=arcsin(3x+1) is -pi/2 <= f(x)<= pi/2

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range of f(x)=arcsin(3x+1)

range

f (x) = arcsin (3 x + 1)

- \frac{π}{2} \leq f (x) \leq \frac{π}{2}

Solution steps

The range of the basic

arcsin

function is

- \frac{π}{2} \leq arcsin (3 x + 1) \leq \frac{π}{2}

- \frac{π}{2} \leq arcsin (3 x + 1) \leq \frac{π}{2}

Therefore the range is

- \frac{π}{2} \leq f (x) \leq \frac{π}{2}

f (x) = \frac{2 x ^{2} + 4 x + 2}{x ^{2} - 1}

f (x) = - x - 3 + 2

y = \frac{5}{2} x, (2, 5)

f (x) = x x

f (x) = x^{4} - 8 x^{2} + 1

What is the range of f(x)=arcsin(3x+1) ?
The range of f(x)=arcsin(3x+1) is -pi/2 <= f(x)<= pi/2