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

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

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inverse of f(x)=3sqrt(x-1)g(x)=x^3+1

inverse

f (x) = 3 x - 1 g (x) = x^{3} + 1

Solution steps

y = x^{3} + 1

Replace

x

with

y

x = y^{3} + 1

Solve for

y, x = y^{3} + 1

f (x) = \frac{1}{12} (- x^{2} + 12 x - 32) + 0.25

f (x) = 2 (y - 3)^{2} + 5

y = - \frac{1}{2} x + 8.5

lo g_{x} (8) = \frac{I}{1 \cdot 1 0 ^{- 12}}

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