Question

Please help!(30pts)(I NEED IT QUICK)Find the inverse of the given function:

Answer 1

Rewrite your function in terms of x and y. Then solve for y in terms of x:

y = -1/2 * (sqrt(x+3)

-2y = sqrt(x+3)

4y^2 = x +3

x = 4y^2-3

Now swap x and y.

y = 4x^2-3

So you got the correct answer. To determine the domain, look back at the original function:

y = -1/2 * (sqrt(x+3), x ≥ -3

We see that x must be greater or equal to -3 in order for the square root to be a real number. Thus sqrt(x+3) must be nonnegative, by that I mean positive or zero.

So if that part of our function is positive and then we are multiplying it by -1/2, our function will only output a nonpositive number, by that I mean either negative or zero. That means our range for this function is y ≤ 0.

And by definition of the inverse, the range of the original function becomes the domain of the inverse. That means the domain of our inverse is x ≤ 0.

I hope that makes sense!

Answer 2

y = -(1/2) sqrt(x + 3) Interchange the x and y

x = -(1/2) sqrt(y + 3) multiply through by - 2

-2x = sqrt(y + 3) Square both sides.

(-2x)^2 = (sqrt(y + 3))^2

4x^2 = y + 3 Subtract 3 from both sides.

y = 4x^2 - 3

The usual rule is to interchange the domain and range of the original equation with the inverse. The range of the original equation is 0 => y > -infinity.

If that is the case, then the domain of the inverse is 0<=x. The graph was revised and provided by Mathmate. It is the correct way to represent this function and it's inverse Including the fact that x<=0. That restriction is very important. The line is x = y and the graphs should reflect around that line.