f(x) = x^2 + 3 with x <or= 0?

Question

a) find the inverse f^-1(x) and state its domain and range.
How can I do it???

Answer 1

y = x^2 + 3

rewrite x = ...
next swap x and y
.
y = x^2 + 3
x^2 = y-3
x = - sqrt(y-3); ( minus since x <= 0 )

swap
y = -sqrt(x-3)
so :
f^-1(x) = - sqrt(x-3)

domain : x>=3
range (-INF,0]

Answer 2

f(x) = x^2 + 3 with x f^-1(x)
[(x^2) + 3]^-x
when x = 0 f(x) = 0
when x< 0 being x negative you can express your eqyaution:
[(x^2) + 3]^x
its a curb situated in the first quadrant (x = positive & y = positive) which limit is infinite.

Answer 3

Switch x and y, and solve for y = f^-1(x),
f^-1(x) = - √(x-3)
D: x ≥ 3
R: y ≤ 0