What is the inverse of f(x)=3(x-1)^2+4 ?

The inverse of f(x)=3(x-1)^2+4 is sqrt((x-4)/3)+1,-sqrt((x-4)/3)+1

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inverse of f(x)=3(x-1)^2+4

inverse

f (x) = 3 (x - 1)^{2} + 4

\frac{x - 4}{3} + 1, - \frac{x - 4}{3} + 1

Solution steps

y = 3 (x - 1)^{2} + 4

Replace

x

with

y

x = 3 (y - 1)^{2} + 4

Solve for

y, x = 3 (y - 1)^{2} + 4

\frac{x - 4}{3} + 1, - \frac{x - 4}{3} + 1

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

\frac{2 x - 5}{x - 7}

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

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

arccos (x)

What is the inverse of f(x)=3(x-1)^2+4 ?
The inverse of f(x)=3(x-1)^2+4 is sqrt((x-4)/3)+1,-sqrt((x-4)/3)+1