What is the inverse of f(x)=((x^3)+1)^{0.5} ?

The inverse of f(x)=((x^3)+1)^{0.5} is cube root of x^2-1

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inverse of f(x)=((x^3)+1)^{0.5}

inverse

f (x) = ((x^{3}) + 1)^{0.5}

3 x^{2} - 1

Solution steps

y = (x^{3} + 1)^{0.5}

Replace

x

with

y

x = (y^{3} + 1)^{0.5}

Solve for

y, x = (y^{3} + 1)^{0.5}

3 x^{2} - 1

f (t) = 3 - 2 cos (3 x + π)

lo g_{10} (- 1.2)

ln ((s^{2} - 1) s^{2})

x^{5} + 2

f (x) = 3 x (x - 2)

What is the inverse of f(x)=((x^3)+1)^{0.5} ?
The inverse of f(x)=((x^3)+1)^{0.5} is cube root of x^2-1