Inverse Functions.

Question

Consider the inverse functions f(x)=x²-2 for (√2 ≤x≤4) and g(x)=√x+2 (all being square rooted) for (0≤x≤14).

 

1. Show that f(g(x)) = x and g(f(x)) = x
2. Find algebraically the coordinates of any points of intersection of the two graphs

Answer 1

f(g(x)) = sqrt(x+2)^2 - 2 = x+2-2 = x

 

g(f(x)) = sqrt (  x^2 - 2 + 2) = sqrt( x^2) = x

 

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

 

(x^2 - 2)^2 = x+2

 

 x^4 - 4x^2 + 4 - x - 2 = 0

 

 x^4 - 4x^2 - x + 2 = 0

 

(x^4 - 4x^2) - x + 2 = 0

 

(x^2)(x^2 - 4) - (x - 2) = 0  <--- factor by grouping

 (x^2) (x + 2)(x-2) - (x-2) = 0

 

(x-2)  { (x^2)(x+2) - 1} = 0

(x-2) { x^3 + 2x^2 - 1} = 0

 

Clearly x-2=0 is a zero, so x=2

 

For

x^3 + 2x^2 - 1 = 0 the possible rational zeros are p=q=p/q={+-1}

x=-1 ---> (-1)^3 + 2(-1)^2 - 1 = -1 + 2 - 1 = 0, so x=-1 is also a zero

 

Synthetic division says:

 

-1 |     1   2  0   -1

              -1  -1   1

        ________________

           1  1   -1  0

 

 x^2 + x - 1 = 0 is the remaining quadratic after x+1 is factored

 

 quadratic formula says:

 

   [-1 +or- sqrt( 1^2 -4(1)(-1))] / 2 =

    [-1 +or- sqrt( 5)]   / 2 are the 2 irrational zeros

 

The negative branch checks out as (-1.618,0.618)

note that 0.618 is approximately the positive branch of these irrational solutions

 

 

x=2 works out as a valid solution since x^2 - 2 = 2^2 - 2 = 4-2 = 2 = sqrt(4) = sqrt(2+2) = sqrt(x+2)

So the inverse functions f and g meet at (2,2)

 

-1 however does not check out as f(-1) = -1 while g(-1) = 1, which occurs as an extraneous zero from squaring both sides

 

 

 

Answer 2

Hi Elouise, I'm sorry my answer had typos. Mark's work is really good.