What is the inverse of f(x)=\sqrt[3]{x-3}+7/5 ?

The inverse of f(x)=\sqrt[3]{x-3}+7/5 is x^3+32/125-(21x^2)/5+(147x)/(25)

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inverse of f(x)=\sqrt[3]{x-3}+7/5

inverse

f (x) = 3 x - 3 + \frac{7}{5}

x^{3} + \frac{32}{125} - \frac{21 x ^{2}}{5} + \frac{147 x}{25}

Solution steps

y = 3 x - 3 + \frac{7}{5}

Replace

x

with

y

x = 3 y - 3 + \frac{7}{5}

Solve for

y, x = 3 y - 3 + \frac{7}{5}

x^{3} + \frac{32}{125} - \frac{21 x ^{2}}{5} + \frac{147 x}{25}

2 x - 3 - 5

3 2^{\frac{2}{5}}

f (x) = 3.174 \cdot x^{2} - 83741 \cdot x + 2771.2

f (x) = 3 + 4 x

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

What is the inverse of f(x)=\sqrt[3]{x-3}+7/5 ?
The inverse of f(x)=\sqrt[3]{x-3}+7/5 is x^3+32/125-(21x^2)/5+(147x)/(25)