What is the inverse of y=(x-3)^2,x<= 3 ?

The inverse of y=(x-3)^2,x<= 3 is sqrt(x)+3,-sqrt(x)+3

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inverse of y=(x-3)^2,x<= 3

inverse

y = (x - 3)^{2}, x \leq 3

x + 3, - x + 3

Solution steps

y = (x - 3)^{2}

Replace

x

with

y

x = (y - 3)^{2}

Solve for

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

x + 3, - x + 3

3 - (- 1)^{n}

f (x) = 16 x^{2} - 56 x + 37

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

lo g_{2} (\frac{- x - 1}{x - 1})

f (x) = \frac{2 x + 1}{5 - 3 x}

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