What is the inverse of f(x)=(sqrt(x^3-1))/4 ?

The inverse of f(x)=(sqrt(x^3-1))/4 is cube root of 16x^2+1

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inverse of f(x)=(sqrt(x^3-1))/4

inverse

f (x) = \frac{x ^{3} - 1}{4}

3 16 x^{2} + 1

Solution steps

y = \frac{x ^{3} - 1}{4}

Replace

x

with

y

x = \frac{y ^{3} - 1}{4}

Solve for

y, x = \frac{y ^{3} - 1}{4}

3 16 x^{2} + 1

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

f (x) = 0.5055 x + 0.0469

\frac{( 9 t ^{2} + 16 ) ^{\frac{3}{2}} - 64}{9}

f (x) = \frac{8}{- 4 x + 1}

y = \frac{2 x ^{2} + 3}{5 - x ^{2}}

What is the inverse of f(x)=(sqrt(x^3-1))/4 ?
The inverse of f(x)=(sqrt(x^3-1))/4 is cube root of 16x^2+1