What is the inverse of f(x)= 2/3 x^3+1 ?

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

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inverse of f(x)= 2/3 x^3+1

inverse

f (x) = \frac{2}{3} x^{3} + 1

Solution steps

y = \frac{2}{3} x^{3} + 1

Replace

x

with

y

x = \frac{2}{3} y^{3} + 1

Solve for

y, x = \frac{2}{3} y^{3} + 1

f (x) = \frac{( x + 3 )}{x + 9}

\frac{- 4 - 3 x}{7 x - 5}

f (x) = x^{3} - 3 x^{2} + 2 x + 1

f (x) = - x^{2}

f (x) = \frac{x ^{2} + x - 2}{x ^{2} - 3 x - 4}

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