# domain of a 1-1 function and its inverse

• Mar 6th 2010, 01:15 PM
break
domain of a 1-1 function and its inverse
Find the domain on which y = x^2 + 8x + 12 is a 1-1 function, and find the inverse function y=f^-1.

how do i find the domain and inverse? thanks!
• Mar 6th 2010, 01:18 PM
Miss
Quote:

Originally Posted by break
Find the domain on which y = x^2 + 8x + 12 is a 1-1 function, and find the inverse function y=f^-1.

how do i find the domain and inverse? thanks!

Its actual domain is R.
To make it a 1-1 function you should let $\displaystyle x \leq -4$ or $\displaystyle x \geq -4$. Do you know why?

To find the inverse function, Let $\displaystyle y=f(x)$ and solve this equation for $\displaystyle x$.
• Mar 6th 2010, 02:08 PM
HallsofIvy
Quote:

Originally Posted by break
Find the domain on which y = x^2 + 8x + 12 is a 1-1 function, and find the inverse function y=f^-1.

how do i find the domain and inverse? thanks!

Since the graph of that function is a parabola, it will be one to one on the two intervals: x< x-coordinate of vertex and x> x-coordinate of vertex.

You can complete the square to find the vertex.

To find the two inverses, write the equation as $\displaystyle x^2+ 8x+ 12-y= 0$ an use the quadratic formula to solve for x.
• Mar 6th 2010, 02:22 PM
drumist
Quote:

Originally Posted by Miss
Its actual domain is R.
To make it a 1-1 function you should let $\displaystyle x \leq 0$ or $\displaystyle x \geq 0$. Do you know why?

To find the inverse function, Let $\displaystyle y=f(x)$ and solve this equation for $\displaystyle x$.

The idea of this is correct, but the correct value to "split" the domain at is not x=0.
• Mar 6th 2010, 02:42 PM
Miss
Quote:

Originally Posted by drumist
The idea of this is correct, but the correct value to "split" the domain at is not x=0.

Oops. Its a typo :o
It shoud be at $\displaystyle x=-4$ :)