Let #f(x) = 1/x^2# and #g(x) = x−1#, how do you find each of the compositions and domain and range?

I will show you how to do #ƒ(g(x))#. This means that we must plug in function g in for x in function ƒ.

#ƒ(g(x)) = ƒ(x - 1)#

#= 1/(x - 1)^2#

#=1/(x^2 - x - x + 1)#

#=1/(x^2 - 2x + 1)#

So, #ƒ(g(x)) = 1/(x^2 - 2x + 1)#

As for the domain and range, the domain are all permissible values of x and the range is the same, except for the values of y.

Here is the graph of the new function.

graph{1/(x^2 - 2x + 1) [-10, 10, -5, 5]}

When a number replaces x in the parentheses (e.g: #ƒ(g(x)) -> ƒ(g(2))#), you must first plug the value in the parentheses into the inner function (g in this case), and then plug in the answer of that calculation into the outer function (ƒ in this case). If you have more than 2 functions, do the same thing, but just remember to work from the inside to the outside.

Since having the denominator equal to 0 in a rational or reciprocal function is undefined (because division by 0 is undefined), we must set the denominator to 0 and solve for x.

#x^2 - 2x + 1 = 0#

#(x - 1)(x - 1) = 0#

#x = 1#

The domain would be #x in RR# | #x != 1#.

Range: Looking at the graph, you can see, as you zoom out, that the graph becomes extremely close to, but never touches, the horizontal line y = 0. Also, you can see that the function never goes below the line y = 0. In other words, the y value must always stay positive.

So, the range is #y > 0#

Practice exercises:

1. Assuming #ƒ(x) = (2x + 6)/2#, #g(x) = sqrt(3x - 5)# and #h(x) = (x + 4)/(5x - 7)#, find:

a) #g(h(x))#

b) #h(ƒ(x))#

c) #g(ƒ(8))#

d) #ƒ(g(h(x)))#

Good luck!