What is the inverse of f(x)=1+\sqrt[3]{2+3x} ?

The inverse of f(x)=1+\sqrt[3]{2+3x} is (x^3-3x^2+3x-3)/3

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inverse of f(x)=1+\sqrt[3]{2+3x}

inverse

f (x) = 1 + 3 2 + 3 x

\frac{x ^{3} - 3 x ^{2} + 3 x - 3}{3}

Solution steps

y = 1 + 3 2 + 3 x

Replace

x

with

y

x = 1 + 3 2 + 3 y

Solve for

y, x = 1 + 3 2 + 3 y

\frac{x ^{3} - 3 x ^{2} + 3 x - 3}{3}

sin (3 0^{\circ})

f (x) = 7 sin (x) - 2

f (x) = 7 sin (x) - 6

f (x) = 7 sin (x) + 1

f (x) = 7 sin (x) + 6

What is the inverse of f(x)=1+\sqrt[3]{2+3x} ?
The inverse of f(x)=1+\sqrt[3]{2+3x} is (x^3-3x^2+3x-3)/3