What is the inverse of f(x)=4\sqrt[3]{x}+1 ?

The inverse of f(x)=4\sqrt[3]{x}+1 is (x^3)/(64)-(3x^2)/(64)+(3x)/(64)-1/64

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inverse of f(x)=4\sqrt[3]{x}+1

inverse

\frac{x ^{3}}{64} - \frac{3 x ^{2}}{64} + \frac{3 x}{64} - \frac{1}{64}

Solution steps

Replace

x

with

y

Solve for

\frac{x ^{3}}{64} - \frac{3 x ^{2}}{64} + \frac{3 x}{64} - \frac{1}{64}

- (7)^{x}

f (x) = \frac{( 2 )}{( x ^{2} - 16 )}

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

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

f (x) = x^{2} - 81

What is the inverse of f(x)=4\sqrt[3]{x}+1 ?
The inverse of f(x)=4\sqrt[3]{x}+1 is (x^3)/(64)-(3x^2)/(64)+(3x)/(64)-1/64