Combination of functions is used to create new functions from original functions. The method of doing so is similar to the method of creating new numbers from the original numbers by using the arithmetic operators. There are four types of combination possible to compose new functions. They are either add, subtract, multiply or divide. So, if we take up two original functions says $f(x)$ and $g(x)$, then the different types of combination of functions would be $f(x) + g(x),\ f(x)  g(x),\ f(x) \times g(x)$ and $\frac{f(x)}{g(x)}$.

Composite functions are formed from combination of functions. In this process two functions are combined such that the first function is performed first and then it is substituted in place of each independent variable '$x$' in the second function and then simplified. Suppose the two functions are $f(x)$ and $g(x)$, then the symbol to denote the composite function would be $(f o g)(x)$ = $f[g(x)]$. Composite functions are not commutative, which mean $f[g(x)]$ is not equal to $g[f(x)]$. The domain of the composite function is the values of the independent variable for which the result given by the first function must lie in the domain of the second function.
Graph of composite function is drawn by first combining the individual functions and then finding out the domain and range of the composite function. The domain and range of the combination of functions gives us the $x$ and $y$ coordinates to plot on the graph. Graphing of composite function is similar to drawing the graph of any ordinary function. The restriction lies that we need to first carry out the combination of function to get the equation of the composite function and then the domain and range is found out to plot on the graph.
Two functions are combined to form a composite function. The domain of the composite function is dependent on the domain of both the first function and the second function. We need to know the domain of both $f(x)$ and $g(x)$ so as to find out the domain of $(f\ o\ g)(x)$. In $f[g(x)]$ we can see that $x$ is a member of the domain of $g$ to make it a valid expression else the inner function $g(x)$ cannot be evaluated. Similarly, we also get to see that $g(x)$ is within the function $f$, so $g(x)$ must be a member of the domain of $f(x)$, else we cannot compute the composite function $f[g(x)]$ evaluation and it make it a valid expression. Thus, combining the individual domains of $f$ and $g$ we can say that the domain of the composite function consist of only those value of inputs in the domain of $g$ which produces the output from $g(x)$ that resides in the domain of $f$. For the composite function $(f\ o\ g)(x)$, the domain would be values of $x$ that can be plugged into $g$ to give us a value of $g(x)$ that can again be substituted into $f$ to solve for $(f\ o\ g)(x)$. Similarly, for the composite function $(g\ o\ f)(x)$, the domain would be values $f x$ that can be plugged into $f$ to give us a value of $f(x)$ that can again be substituted into $g$ to solve for $(g\ o\ f)(x)$.
Steps to find out the domain of the combination of two functions, namely $f(x)$ and $g(x)$:
Step 1: We need to find the domain of the second function $g(x)$
Step 2: We need to find out the domain of the first function $f(x)$
Step 3: We need to find out only those values of $x$ which belongs to the domain of g for which values of $g(x)$ computed lies in the domain of $f$. We need to exclude those values of $x$ which does not lie in the domain of $g$ for which $g(x)$ does not lie in the domain of $f$
Step 4: The resulting set $f$ values gives us the domain of the composite function $(f\ o\ g)(x)$
There are four types of combinations of composite functions which can be carried out. Like we do with the numbers, similarly while combining functions we apply arithmetic combinations. The arithmetic combinations of functions are sum, difference, product or quotient of functions. Let us take up two individual functions $f(x)$ and $g(x)$, then the arithmetic combinations of functions are:
Sum: $(f + g)(x)$ = $f(x) + g(x)$
Where, $(f + g)(x)$ is the composite function
Difference: $(f  g)(x)$ = $f(x)  g(x)$
Where, $(f  g)(x)$ is the composite function
Product: $(f \times g)(x)$ = $f(x) \times g(x)$
Where, $(f \times g)(x)$ is the composite function
Quotient: $(\frac{f}{g})$ $(x)$ = $\frac{f(x)}{g(x)}$
Where, $(\frac{f}{g})$ $(x)$ is the composite function
Point to remember for the quotient rule the denominator $g(x)$ cannot be equal to $0$, else the composite function $(\frac{f}{g})$ $(x)$ would become invalid.
Examples 1:
Given $f(x)$ = $x + 4$ and $g(x)$ = $5  x^2$, find the following :
a) $(f\ o\ g)(x)$
b) $(g\ o\ f)(x)$
c) $(f\ o\ g)(2)$
d) $(g\ o\ f)(1)$
Solution:
a) The composition of the function $f$ with the function $g$ is
$(f\ o\ g)(x)$ = $f[g(x)]$ = $g(x) + 4$ = $5  x^2 + 4$ = $9  x^2$
b) The composition of function $g$ with the function $f$ is
$(g\ o\ f)(x)$ = $g[f(x)]$ = $5  f(x)^2$ = $5  (x + 4)^2$ = $5  (x^2 + 8x +16)$ = $5  x^2  8x  16$ = $ x^2  8x  11$
c) We need to plugin the value of x =2 in the composite function $(f\ o\ g)(x)$ to find out $(f\ o\ g)(2)$
$(f\ o\ g)(2)$ = $9  2^2$ = $9  4$ = $5$
d) We need to substitute the value of $x$ = $1$ in the composite function $(g\ o\ f)(x)$ to find out $(g\ o\ f)(1)$
$(g\ o\ f)(1)$ = $ 1^2  8 \times 1  11$ = $1  8  11$ = $20$
Example 2:
A refrigerated food has number of bacteria $(N)$ given by $N(T)$ = $10T^2  40T + 400$, where $T$ stands for the temperature of the food measured in degree celsius and the range of value $T$ lies is $2 \Leftarrow T \Leftarrow 10$. The food is then taken out from the refrigerator and the temperature of the food rises and is given by $T(t)$ = $2t + 5,\ 0 \Leftarrow t \Leftarrow 4$, where $t$ stands for the time for which it is kept out of the refrigerator in hours
a) Find out the composition of the function $N(T)$ with the function $T(t)$, also interpret its result
b) Find out the time when the bacteria count reaches $1500$
Solution:
a) The composition of the function $N(T)$ with the function $T(t)$ is given by
$N[T(t)]$ = $10T(t)^2  40T(t) + 400$ = $10(2t+ 5)^2  40(2t + 5) + 400$
$N[T(t)]$ = $10(4t^2 + 20t + 25)  40(2t + 5) + 400$
$N[T(t)]$ = $40t^2 + 200t + 250  80t  200 + 400$
$N[T(t)]$ = $40t^2 + 120t + 450$
The combination of functions $N(T)$ and $T(t)$ which is $N[T(t)]$ denotes the number of bacteria residing in the food as a function of the amount of time for which the food has been taken out of the refrigerator.
b) The bacteria count will reach $1500$ when the composite function
$N[T(t)]$ = $1500$
$40t^2 + 120t + 450$ = $1500$
$40t^2 + 120t + 450  1500$ = $0$
$40t^2 + 120t  1050$ = $0$
$4t^2 + 12t  105$ = $0$
Applying the quadratic formula to solve for $t$,
$'t$ = $ 12 \pm$ $\frac{\sqrt{(12^2  4 \times 4 \times 105)}}{2 \times 4}$
$'t$ = $\frac{12 \pm \sqrt{1824}}{8}$
$'t$ = $\frac{12 + 42.71}{8}$
$'t$ = $\frac{30.71}{8}$
$'t$ = $3.84$ hours
Thus, after $3.84$ hours the bacteria count shall reach $1500$
Example 3:
Find the domain and range of the following function:
a) $f(x)$ = $x + 1$
b) $g(x)$ = $\sqrt{x  5}$
c) $(f\ o\ g)(x)$
d) $(g\ o\ f)(x)$
Solution:
a) For function $f(x)$ = $x + 1$ domain is all real numbers and range is also all real numbers
b) For function $g(x)$ = $\sqrt{x  5}$ domain shall be $x \Rightarrow 5$ and range shall be $\sqrt{x  5 \Rightarrow 0}$, so $g(x) \Rightarrow 0$
c) $(f\ o\ g)(x)$ = $f[g(x)]$ = $g(x) + 1$
$f[g(x)]$ = $\sqrt{(x  5)}\ +\ 1$
Domain of $f[g(x)]$ = $\sqrt{(x  5)} + 1$ is $x \Rightarrow 5$
Range of $f[g(x)]$ = $\sqrt{(x  5)} + 1 \Rightarrow$ Since $\sqrt{x  5} \Rightarrow 0$, $\sqrt{x  5} + 1 \Rightarrow 1$, so range of $f[g(x)] > 1$
d) $(g\ o\ f)(x)$ = $g[f(x)]$ = $\sqrt{f(x)  5}$ = $\sqrt{x + 1  5}$ = $\sqrt{x  4}$
Domain of $(g\ o\ f)(x)$ = $\sqrt{x  4}$ is $\sqrt{x  4} \Rightarrow 0$, so $x \Rightarrow 4$
Range of $(g\ o\ f)(x)$ = $\sqrt{x  4}\ \Rightarrow$ Since $\sqrt{x  4}\ \Rightarrow 0$, so range of $g[f(x)] \Rightarrow 0$