The one-to-one functions:

Question

The one-to-one functions g and h are defined as follows:

 

g= {(2, 0) (−5, 4) (6, 5) (−8, 9)

h(x)=3x+8

 

Find the Following:   g^-1(5)=

 

                             h^-1(x)=

                             

                             (h^-1 o h)(-4)=

h 

=hx+3x8

 

Answer 1

A function is a mapping from one set of objects D (called the domain) onto another set of objects R (called the range).  Functions can be represented by ordered pairs (x,y) which means that the function maps the object x (which must be an element of the set D) to the object y (which must be an element of the set R).  The inverse function simply reverses this process and maps the object y (from the set R) back to the object x (in the set D), so the inverse function is represented by the ordered pairs (y,x).  Since the function g contains the ordered pair (6,5) (which means that g maps the element 6 from D to the element 5 in R), the inverse function g^-1 will map the element 5 (from R) back to the element 6 (in D), which is represented by the ordered pair (5,6), which means that g^-1(5) = 6.  

 

If you are given an equation for a function you can find the inverse function by the following procedure  

Let y = h(x) so  y = 3x + 8

Switch x and y to get  x = 3y + 8

Solve this equation for y to get  y = (x-8)/3 = (1/3)x - 8/3  

y is the inverse function so  h^-1(x) = (1/3)x - 8/3  

 

The function composition (h^-1 o h)(-4) means h^-1(h(-4)).  This means that you first evaluate h(-4) and then plug whatever value you get into h^-1 and evaluate h^-1 at that value.  By the definition of inverse function in the 1st paragraph above, h^-1 must map this value (whatever it is) right back to (-4).  So it must be the case that h^-1(h(-4)) = -4.  If you are curious you can plug (-4) into h to get h(-4) = 3*(-4) + 8 = -12 + 8 = -4, and then plug this value (which coincidentally happens to be -4, which will not always be the case) into h^-1 to get h^-1(-4) = (1/3)*(-4) - 8/3 = -4/3 - 8/3 = -12/3 = -4