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

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

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

inverse

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

\frac{x - 1}{2} + 3, - \frac{x - 1}{2} + 3

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x - 1}{2} + 3, - \frac{x - 1}{2} + 3

f (x) = e^{4 x + 4}

f (x) = 11 x + 12

\frac{x + 7}{x - 4}

f (x) = - 8 - x

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

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