SOLUTION: Suppose H(x)=√6x-1. Find two functions f and g such that (f °g)(x) = H(x) f(x) = g(x) = I don't understand how I figure out the functions.

Algebra ->  Functions -> SOLUTION: Suppose H(x)=√6x-1. Find two functions f and g such that (f °g)(x) = H(x) f(x) = g(x) = I don't understand how I figure out the functions.       Log On


   



Question 1168502: Suppose H(x)=√6x-1.
Find two functions f and g such that
(f °g)(x) = H(x)
f(x) =
g(x) =
I don't understand how I figure out the functions.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


If you use the "√" symbol in your post, use parentheses to make the meaning clear.

Is H(x) equal to √6x-1 = sqrt%286x%29-1?
Or is it equal to √(6x-1) = sqrt%286x-1%29?

I suspect it is supposed to be the second; but it could be either one.

I will go ahead and explain how to answer the question if H%28x%29+=+sqrt%286x%29-1.

If it is the other one, you can use my discussion below to answer the question for that other case.

Look at the function and see what it does to the input value x:
(1) multiply by 6;
(2) take the square root; and
(3) subtract 1

To make the function H(x) a composition of two function f(x) and g(x), simply condense that sequence of three operations into a sequence of two operations, keeping all the operations in the original order.

There are two possibilities:

One choice is to combine steps (1) and (2):
(1) multiply by 6 and take the square root; and
(2) subtract 1

Then the function g(x) is the first of those steps: g%28x%29+=+sqrt%286x%29;
and the function f(x) is the second: f%28x%29+=+x-1

ANSWER (one possibility): f%28x%29+=+x-1 and g%28x%29+=+sqrt%286x%29

The other choice is to keep step 1 by itself and combine steps 2 and 3:
(1) multiply by 6; and
(2) take the square root and subtract 1

Then the function g(x) is g%28x%29+=+6x;
and the function f(x) is f%28x%29+=+sqrt%28x%29-1

ANSWER (a second possibility): f%28x%29+=+sqrt%28x%29-1 and g%28x%29+=+6x