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

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

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

inverse

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

x - 4 - 1, - x - 4 - 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

x - 4 - 1, - x - 4 - 1

in v erse f (x) = \frac{5 x}{x - 6}

in v erse 3 x^{3}

cr i t i c a l y = \frac{2 x ^{2} - 5 x + 5}{x - 2}

in v erse f (x) = 4 - x^{2}, x \geq 0

in v erse f (x) = 3 - 2 y^{3}

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