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

The inverse of f(x)=sqrt((x-3)/3)+1 is 3x^2-6x+6

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

inverse

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

3 x^{2} - 6 x + 6

Solution steps

y = \frac{x - 3}{3} + 1

Replace

x

with

y

x = \frac{y - 3}{3} + 1

Solve for

y, x = \frac{y - 3}{3} + 1

3 x^{2} - 6 x + 6

f (x) = \frac{x ^{2} + 4 x - 5}{x ^{2} - 9}

x^{2} - 1 x = 1

- 2 sin (x)

f (x) = x^{2} - 1

f (x) = 5 x^{2} + 2 x - 1

What is the inverse of f(x)=sqrt((x-3)/3)+1 ?
The inverse of f(x)=sqrt((x-3)/3)+1 is 3x^2-6x+6