Help me solve this function.?

Question

x = y^2 + 2y + 5

Domain=

Range=

Answer 1

Since "domain" and "range" can be interpreted differently depending on what you assume your variable is, I will give you the set of values of x such that there is a y that solves the equation x = y^2 + 2y +5. Then I'll do the same for y.

Note that y^2 + 2y + 5 = (y + 1)^2 + 4 >=4 and, for y=-1, we have

(-1)^2 + 2*(-1) +5 = 4. Thus, if x < 4, there is no solution to the equation.

We've already seen that x = 4 and y= - 1 solve the equation, so x can be equal to 4.

Assume x > 4. Then the equation is equivalent to

(y + 1)^2 = x - 4

since x > 4, it follows that x - 4 > 0, and thus y = -1 + sqrt(x - 4), together with x > 4 is a solution.

Hence, for any x >=4 there is a y that solves the equation. We conclude that

x is in [4,+ infinity)

Now note that, for a given y, the expression y^2 + 2y + 5 is well defined, and thus there exist a number, call it x, such that

x = y^2 + 2y + 5. We conclude that

y is in (- infinity, + infinity)

Hope this helps.