Help finding domain and inverse function?

y = 5/(x + 3) The denominator can't be zero x can be anything except -3.so x=-3 would be the vertical asymptone and the domain woul be all reals exept for -3..........(-infinity,-3) (-3,infinity) . and to find the inverse oparation just substitude x for y and y for x and solve it. g(x)=square root x----->y=square root x x=square root y x^2=y this would be the inverse

What I always do is draw out the function: Take everything you know. If there's x+3 in the denominator, then 5/0 is possible when x = -3. 5/0 is not a real number, so you know there's an asymptote at x = -3. You know that 5 divided by a negative number can give you a real y-coordinate, and a positive number can also, so the only number that won't work is -3. Therefore: x ≠ 3. ---------------------- Now for the second part: Change g(x) to y and rearrange to make x the subject. y = √x x = y² Now swap the variables: y = x², so inverse(g(x)) = x²

y = 5/(x + 3) The denominator can't be zero: x + 3 <> 0 x <> -3 x can be anything except -3. g = sqrt(x) Square both sides: g^2 = x To find the inverse, swap the variables: x^2 = g g = x^2

What I always do is draw out the function: Take everything you know. If there's x+3 in the denominator, then 5/0 is possible when x = -3. 5/0 is not a real number, so you know there's an asymptote at x = -3. You know that 5 divided by a negative number can give you a real y-coordinate, and a positive number can also, so the only number that won't work is -3. Therefore: x ≠ 3. ---------------------- Now for the second part: Change g(x) to y and rearrange to make x the subject. y = √x x = y² Now swap the variables: y = x², so inverse(g(x)) = x²

y = 5/(x + 3) The denominator can't be zero: x + 3 <> 0 x <> -3 x can be anything except -3. g = sqrt(x) Square both sides: g^2 = x To find the inverse, swap the variables: x^2 = g g = x^2