What is the inverse and domain of f (x) = x^2+4

To find the inverse let

y = x^2 + 4

x^2 = y - 4

x = sqrt (y - 4)

inverse f-1 (x) = sqrt (x - 4)

the domain of x^2 + 4 is all real x

the domain of f-1 (x) is x> = 4

The domain of the inverse of a function is the range of the original function

so find the range of f (x)

this is lowest to highest

we got the vertex is at (0,4)

and it goes to infinity

so range is from 4 to infinity

inverse

replace f (x) with y

switch x and y

solve for y

replace y with finverse

f (x) = x^2+4

y=x^2+4

x=y^2+4

x-4=y^2

y=√ (x-4)

f⁻¹ (x) = √ (x-4)

inverse is f⁻¹ (x) = √ (x-4)

the domain is x≥4