What is the inverse of f(x)= 1/7 (x^3-1) ?

The inverse of f(x)= 1/7 (x^3-1) is cube root of 7x+1

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inverse of f(x)= 1/7 (x^3-1)

inverse

f (x) = \frac{1}{7} (x^{3} - 1)

Solution steps

y = \frac{1}{7} (x^{3} - 1)

Replace

x

with

y

x = \frac{1}{7} (y^{3} - 1)

Solve for

y, x = \frac{1}{7} (y^{3} - 1)

\frac{x + 6}{x - 2}

36 - x^{2}

(- 3, - 4), (5, - 2)

f (x) = \frac{( x + 7 )}{( x - 6 )}

f (x) = x + 1

What is the inverse of f(x)= 1/7 (x^3-1) ?
The inverse of f(x)= 1/7 (x^3-1) is cube root of 7x+1