What is the inverse of f(x)= 1/7 (x^3+5) ?

The inverse of f(x)= 1/7 (x^3+5) is cube root of 7x-5

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inverse of f(x)= 1/7 (x^3+5)

inverse

f (x) = \frac{1}{7} (x^{3} + 5)

Solution steps

y = \frac{1}{7} (x^{3} + 5)

Replace

x

with

y

x = \frac{1}{7} (y^{3} + 5)

Solve for

y, x = \frac{1}{7} (y^{3} + 5)

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

y = x^{2} + 9, x \leq 0

- x^{2} + 2 x + 5

f (x) = \frac{1}{2 x}, x \neq = 0

f (x) = 13 x + 1

What is the inverse of f(x)= 1/7 (x^3+5) ?
The inverse of f(x)= 1/7 (x^3+5) is cube root of 7x-5