What is the inverse of f(x)={x^3+1,x<= 0} ?

The inverse of f(x)={x^3+1,x<= 0} is cube root of x-1

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inverse of f(x)={x^3+1,x<= 0}

inverse

f (x) = {x^{3} + 1, x \leq 0}

Solution steps

y = x^{3} + 1

Replace

x

with

y

x = y^{3} + 1

Solve for

y, x = y^{3} + 1

(1, - 5), (- 7, 1)

f (x) = - x^{3} + 5 x + 9

f (x) = \frac{5 + x}{9 - x}

f (x) = (\frac{- x ^{4}}{4}) + x^{2} - 1

(4, 5), (3, 2)

What is the inverse of f(x)={x^3+1,x<= 0} ?
The inverse of f(x)={x^3+1,x<= 0} is cube root of x-1