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

The inverse of f(x)=2(x^{1/3}+1) is (x^3)/8-(3x^2)/4+(3x)/2-1

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

inverse

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

\frac{x ^{3}}{8} - \frac{3 x ^{2}}{4} + \frac{3 x}{2} - 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x ^{3}}{8} - \frac{3 x ^{2}}{4} + \frac{3 x}{2} - 1

f (x) = x^{3} - 3 x + 6

y = - 2 x - 4

x^{\frac{2}{3}} (8 - x)

f (x) = \frac{3 x + 9}{1 - 2 x}

f (x) = 0.15 (x - 3000) + 300

What is the inverse of f(x)=2(x^{1/3}+1) ?
The inverse of f(x)=2(x^{1/3}+1) is (x^3)/8-(3x^2)/4+(3x)/2-1