Does anyone know how to find the domain and range of this inverse function?

Question

I keep getting the wrong awnser thanks for the help

f(x)=square root(2x+5)
f inverse (x)= 1/2x^2 - 5/2

Answer 1

To clarify/slightly tweak another answer above,

for the inverse function:
domain is all real numbers,
and range is greater than OR equal to -5/2

If x is 0, y is -5/2, and if if x is any other real number, y is greater than -5/2, hence the range.

As far as a domain, any real number will give you a real number. X can be any real number, positive or negative, since x^2 will always yield a positive number.

These are also the inverses of the domain and range of the original function.

Answer 2

At first 2x+5>=0 So x>= -5/2 y= sqrt(2x+5) is positive or zero
x= 1/2(y^2-5) So as inverse function and if you wish to interchange x and y

y=1/2(x^2-5) with domain x>=0 and range y>=-5/2

Answer 3

For the inverse function, the domain is all real x, and the range is all values greater than -5/2

Answer 4

f(x) domain x >= -5/2 range y >=0

inverse domain all reals range >= sq rt 5