# Operating Composite Functions

In the previous posts I gave overviews about domain and range as well as how to determine whether an equation is a function or not. This time let’s talk about operating composite functions. Composite functions are combination of two or more functions.

Worked Example 1:

Given $g(x)=2x+1$ , $h(x)=x^2-1$ What is h◦g?

Solution:

h◦g means h that is compose of g(x)

h◦g = h(g(x))

$h(x)=x^2-1$ $h(g(x))=(2x+1)^2-1$ $h(g(x))=4x^2+4x+1-1$ $h(g(x))=4x^2+4x$

Worked Problem 2: (Metrobank -MTAP Dep-Ed Math Challenge Elimination)

If $f(x)=1-x$ and $g(x)=x^2-1$, what is f◦g(2)?

Solution:

f◦g(x) = f(g(x))

$latex f(x) = 1-x$

$f(g(x)) = 1-(x^2-1)$ $f(g(x))=2-x^2$ $f(g(2))=2-(2)^2$ $f(g(2))=-2$

Worked Problem 3: MMC Elimination 2013

If $f(\frac{x-3}{5-x})=3x-2$, find f(x)

Solution:

Let $g(x)=\displaystyle\frac{x-3}{5-x}$. This will reduce the relation to $f(g(x))=3x-2$

Let’s start with $g(x)=\displaystyle\frac{x-3}{5-x}$ , We need to find the value of x here to make it only x. Meaning, we will look for a specific expression in x to substitute. We do trial and error. Let x=x+1,etc. But of course that is so impractical. Try this out.

Let y=g(x)

$y=\displaystyle\frac{x-3}{5-x}$

We solve for x in terms of y. Why? Because that is the value of x that will make the expression   $\displaystyle\frac{x-3}{5-x}$   become x.

$y=\displaystyle\frac{x-3}{5-x}$ $y(5-x)=x-3$ $5y-xy=x-3$ $5y+3=x+xy$ $5y+3=x(1+y)$ $x=\displaystyle\frac{5y+3}{1+y}$

But that is not the answer yet,

Since we are just working with x’s we need to drop all y’s and put x instead.

$x=\displaystyle\frac{5x+3}{1+x}$

Going back to the original problem,

$f(\displaystyle\frac{x-3}{5-x})=3x-2$

Now, let $x=\displaystyle\frac{5x+3}{1+x}$

$f(\displaystyle\frac{(\displaystyle\frac{5x+3}{1+x})-3}{5-(\displaystyle\frac{5x+3}{1+x})})=3(\displaystyle\frac{5x+3}{1+x})-2$

Manipulating this we get

$f(\displaystyle\frac{2x}{x})=\displaystyle\frac{15x+9-2(1+x)}{1+x}$ $f(x)=\displaystyle\frac{13x+7}{x+1}$

This is the required answer. But if the problem ask for $f(x+1),f(x^2),f(x^2-1)$ etc. From f(x) you can let x=x+1 and so on.

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

### 1 Response

1. April 19, 2014

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