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

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

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

inverse

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

x - 1 + 3, - x - 1 + 3

Solution steps

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

Replace

x

with

y

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

Solve for

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

x - 1 + 3, - x - 1 + 3

f (x) = \frac{x ^{2} + 5}{x ^{2} - 2 x - 15}

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

f (x) = \frac{3 - x}{x + 1} - \frac{4}{5 x - 3}

x^{4} + \frac{4}{3} x^{3} - 4 x^{2} - \frac{4}{3}

\frac{6 x}{1 - 7 x}

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