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

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

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

inverse

(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

f (x) = - 7 x - 3

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

f (x) = 2 lo g_{10} (4 x - 7) + 1

arccos (3 + 2 x)

\frac{1}{x ^{2} - 4 x - 5}

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