Question

How do you find the inverse of `g(x) = x^2 + 4x + 3` and is it a function?

Answer 1

See below.

Explanation:

To find the inverse function, we need to express `x` as a function of `y`:

`y=x^2+4x+3`

Substitute:

`y=x`

`x=y^2+4y+3`

Subtract `x`:

`y^2+4y+3-x=0`

Using the quadratic formula:

`y=(-(4)+-sqrt((4)^2-4(1)(3-x)))/(2(1))`

`y=(-4+-sqrt(16-12+4x))/2`

`y=(-4+-sqrt(4+4x))/2`

`y=(-4+-sqrt(4(1+x)))/2`

`y=(-4+-2sqrt((1+x)))/2=-2+-sqrt((1+x))`

`:.`

`f^-1(x)=-2+sqrt(1+x)`

`f^-1(x)=-2-sqrt(1+x)`

If we look at the inverses, remembering that `x` is the range of the function, we can see that, for:

`sqrt(1+x)`

`1+x>=0`

`x>=-1`

This means that:

`x^2+4x+3>=-1`

Solving:

`x^2+4x+4>=0`

Factor:

`(x+2)^2>=0`

So the inverses are defined for the domain of the function.

They are both functions if we take `sqrt(1+x)` as meaning the principal root.