What is the inverse of f(x)=\sqrt[3]{7-x}+1 ?

The inverse of f(x)=\sqrt[3]{7-x}+1 is -x^3+3x^2-3x+8

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inverse of f(x)=\sqrt[3]{7-x}+1

inverse

- x^{3} + 3 x^{2} - 3 x + 8

Solution steps

Replace

x

with

y

Solve for

- x^{3} + 3 x^{2} - 3 x + 8

4 x - 3

y = \frac{3}{4} x - \frac{3}{2}

\frac{( 2 x - 1 )}{( 5 x - 3 )}

f (x) = x^{2} - 16, x > 0

What is the inverse of f(x)=\sqrt[3]{7-x}+1 ?
The inverse of f(x)=\sqrt[3]{7-x}+1 is -x^3+3x^2-3x+8