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}

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}

f (x) = \frac{22 x - 14}{1}

f (x) = 10 x^{2} - 2

f (x) = 9.45

f (x) = 18 x - 56

1 x + 3

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