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.

composition of function


Worked Example 1:

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


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)?


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)


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)


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.


Going back to the original problem,


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


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.

You may also like...

1 Response

  1. April 19, 2014

    camiseta espa帽a 2014

    I really like what you guys are up too. This type of clever work and coverage! Keep up the terrific works guys I’ve included you guys to my own blogroll.|

Leave a Reply

Your email address will not be published.

Protected by WP Anti Spam