Question 227505
Working with functions becomes much easier if you get a better grasp of what the function rule means and what the role of "x" is in the function rule:<ul><li>The function rule is more than a formula. It is a description, in the language of Mathematics, of how the function works. It describes what the particular function will do with its input in order to come up with its output.</li><li>The role of the "x" in the function rule is as a place-holder. In the parentheses on the left it represents whatever input is provided and on the right it represents whatever the function will do with the input.</li></ul>
Let's apply these ideas to your {{{f(x) = x^2-1}}}. A literal reading of this: The function f squares x and then subtracts one. A better, more enlightened reading: The function f will take its input, whatever it is, squares it and then subtracts 1. If you give f a 10 as input, f will square it and then subtract 1. If you give f a "q", it will square it and then subtract 1. If you give it a {{{3t^2-6t+11}}} as input, f will square it and then subtract 1. No matter what you provide f as imput, f will square it and then subtract 1. This is what the function rule is really saying.<br>
f(g(x)) means make g(x) the input to f. And what will f do?? It will square it and then subtract 1!!:
f(g(x)) = (g(x))^2 -1}}}
Substituting the function rule for g(x) we get:
f(g(x)) = ((1/x)-1)^2 - 1}}}
Your teacher probably wants this simplified. So using FOIL or the pattern {{{(a-b)^2 = a^2 - 2ab + b^2}}} to multiply we get:
{{{f(g(x)) = (1/x)^2 - 2(1/x)(1) + 1^2 - 1}}}
{{{f(g(x)) = 1/x^2 - (2/x) + 1 - 1}}}
{{{f(g(x)) = 1/x^2 - 2/x}}}<br>
Now we can turn our attention to g(f(x)). This says make f(x) the input to the function g. The function g, as we can see from the rule: {{{g(x) = 1/x - 1}}}, will find the reciprocal (1/x is the reciprocal of x) of the input and then subtract 1. So when you give it f(x), it will find the reciprocal of f(x) and then subtract 1:
{{{g(f(x)) = 1/(f(x)) - 1}}}
Substituting the rule for f(x) we get:
{{{g(f(x)) = 1/(x^2-1) - 1}}}
This could be considered simplified enough. Or perhaps we should subtract (by getting common denominators first, of course):
{{{g(f(x)) = 1/(x^2-1) - 1((x^2-1)/(x^2-1))}}}
{{{g(f(x)) = 1/(x^2-1) - (x^2-1)/(x^2-1)}}}
Now we can subtract:
{{{g(f(x)) = (1 - (x^2-1))/(x^2-1)}}}
{{{g(f(x)) = (2 - x^2)/(x^2-1)}}}