What's the inverse of the function: y = 3x^2 + 3?

y = 3x^2 + 3

Flip the x and y to get the inverse:

x = 3y^2 + 3

Solve for y:

(x - 3) = 3y^2

(x-3) / 3 = y^2

y = +/- sqrt((x-3)/3)

y = +/- (sqrt(x-3) / sqrt(3))

In math, it's customary to remove radicals from the denominator. Do this by multiplying top and bottom by the radical in the denominator:

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

y = +/- sqrt(3x - 9) / 3

y = 3x^2 + 3
x = 3y^2 +3
x-3 = 3y^2
y^2=(x-3)/3
y = +/- sqrt[ (x-3)/3]
f^-1(x) = +/- sqrt[ (x-3)/3]
Strictly speaking, the inverse is not a function because there are two values in the range set for one value in the domain. The function fails the vertical line test.

The function, defined for all reals, has no inverse function since the function is not one to one.

If you restrict the domain, then the inverse depends on which domain you choose for the function.

If you go through the usual procedure, reverse x and y then solve for y, you get:

x = 3y^2 + 3

3y^2 = x - 3

y^2 = (x-3)/3

Now if the original domain is restricted to x greater than or equal to 0, then the range of the inverse will be y greater than or equal to 0, and you would take the positive square root:

y = sqrt{(x-3)/3}

but if the domain of the original function is restricted to x less than or equal to 0, then the range of the inverse will be y less than or equal to 0, and you would take the negative square root:

y = - sqrt{(x-3)/3}

In both cases, the domain of the inverse is the range ( or codomain, depending on what term your book uses) of the original function, namely numbers greater than or equal to 3.

Of course, if all you are looking for is an inverse relation, rather than an inverse function, then you do not have to restrict the domain , and then you have "+ or -" in front of the radical sign.

y = 3x^2 + 3
3x^2 = y - 3
x^2 = (y - 3)/3
x = √(y/3 - 3/3)

Change x into f^-1(x) and y into x:
x = √(y/3 - 1)
f^-1(x) = √(x/3 - 1)