What is the inverse of f(x)=(x-3)^2+7,x>3 ?

The inverse of f(x)=(x-3)^2+7,x>3 is sqrt(x-7)+3,-sqrt(x-7)+3

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inverse of f(x)=(x-3)^2+7,x>3

inverse

f (x) = (x - 3)^{2} + 7, x > 3

x - 7 + 3, - x - 7 + 3

Solution steps

y = (x - 3)^{2} + 7

Replace

x

with

y

x = (y - 3)^{2} + 7

Solve for

y, x = (y - 3)^{2} + 7

x - 7 + 3, - x - 7 + 3

y = \frac{1}{40} ln (0.1 x) + 30

- 4 x^{2} - 6 x + 3

f (x) = - e^{- x} + 8

f (x) = - e^{- x} + 5

y = \frac{- x - 13}{7}

What is the inverse of f(x)=(x-3)^2+7,x>3 ?
The inverse of f(x)=(x-3)^2+7,x>3 is sqrt(x-7)+3,-sqrt(x-7)+3