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

The inverse of f(x)=(x/4-(1/4))^{1/3} is 4x^3+1

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

inverse

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

4 x^{3} + 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

4 x^{3} + 1

\frac{2}{5}

f (x) = \frac{x ^{2} + 11 x + 28}{3 x + 12}

(\frac{1}{4}, - \frac{1}{2}), (\frac{3}{4}, 1)

cos^{2} (x)

\frac{5}{4 + x}

What is the inverse of f(x)=(x/4-(1/4))^{1/3} ?
The inverse of f(x)=(x/4-(1/4))^{1/3} is 4x^3+1