domain of the inverse

Question

If f(x) = x^2 + 4x, find (1) a domain on which f(x) is one-to-one, (2) the inverse of f(x) on the domain given in part (1) and (3) the domain of the inverse in part (2).

 

I got (1) domain  is x>= -2 (vertex point)

 

(2) the inverse is -2 + sqrt(x+4)

 

(3) here is where I am confused. Is the domain of the inverse x >= -4 or is it the same domain as f(x), then x>= -2?

 

Thanks!

Answer 1

f(x) = x²+4x = x(x+4)

 

The graph of f is a parabola opening upward with x-intercepts when       x = 0 and when x = -4.  The x coordinate of the vertex lies halfway between the x coordinates of the x intercepts.  So, the vertex is

(-2, f(-2)) = (-2, -4).

 

The function is not 1-1 since a parabola fails the horizontal line test.  If we restrict the domain to x≥-2 (the right half of the parabola), then that function is 1-1 and thus has an inverse function.  

 

f(x) = x² + 4x , x≥-2 has domain [-2,∞) and range [-4, ∞).

 

The domain of f^-1(x) is the same as the range of f(x) and the range of 

f^-1(x) is the same as the domain of f(x).  So, the domain of f^-1(x) is [-4,∞).  

 

 

Answer 2

It is what you thought, x>=4.