Hello everyone! Welcome to this deep dive session on one of the most fundamental yet powerful concepts in functions:
Composition of Functions. This is a concept that forms the backbone for many advanced topics in calculus and beyond, and it's a guaranteed area for JEE Main and Advanced questions, especially concerning domains and ranges.
We'll start from the very basics, build intuition, explore formal definitions, and then tackle advanced examples and JEE-specific considerations.
### What is Composition of Functions?
Imagine you have two machines. The first machine takes an input, processes it, and gives an output. The second machine then takes *that output* as its input, processes it, and gives a final output. This sequential processing, where the output of one function becomes the input for another, is precisely what we call
composition of functions.
Think of it like an assembly line:
1.
Machine 1 (Function `f`): Takes raw material (an input `x`), processes it, and produces an intermediate product (`f(x)`).
2.
Machine 2 (Function `g`): Takes that intermediate product (`f(x)`), processes it further, and produces a final finished product (`g(f(x))`).
So, in essence, composition is about chaining functions together. We are applying one function, and then applying another function to the result of the first.
### Formal Definition of Composition
Let's get formal. Suppose we have two functions:
* A function $f: A o B$ which maps elements from set $A$ to set $B$. So, for any $x in A$, $f(x) in B$.
* A function $g: B o C$ which maps elements from set $B$ to set $C$. So, for any $y in B$, $g(y) in C$.
For the composition of $f$ and $g$ to be defined, specifically $g circ f$ (read as "g of f" or "g composed with f"), a crucial condition must be met:
The
range of the inner function ($f$) must be a subset of the domain of the outer function ($g$).
In simpler terms, every output produced by $f$ must be an acceptable input for $g$.
If this condition holds, we can define a new function, $g circ f$, as follows:
The composite function $(g circ f): A o C$ is defined by:
$$ mathbf{(g circ f)(x) = g(f(x))} quad ext{for all } x in A $$
Here, you first apply $f$ to $x$ to get $f(x)$, and then you apply $g$ to the result $f(x)$.
Key Insight: Notice the order. Even though it's written $g circ f$, the function $f$ is applied first, then $g$. Think of it as reading from right to left in $g(f(x))$.
What about $f circ g$?
If we have $g: A o B$ and $f: B o C$, then $(f circ g)(x) = f(g(x))$ provided the range of $g$ is a subset of the domain of $f$.
Critical Check for JEE: Before attempting to find the expression for a composite function or its domain, always check if the composition is even *defined*. The range of the inner function must be a subset of the domain of the outer function. If this condition is not met, the composite function is undefined for some or all values.
Let's summarize the domains and ranges:
Function |
Domain |
Range |
|---|
$f$ |
$D_f$ |
$R_f$ |
$g$ |
$D_g$ |
$R_g$ |
$g circ f$ |
${x in D_f mid f(x) in D_g}$ |
${g(y) mid y in R_f ext{ and } y in D_g}$ |
$f circ g$ |
${x in D_g mid g(x) in D_f}$ |
${f(y) mid y in R_g ext{ and } y in D_f}$ |
### Properties of Composition of Functions
Understanding these properties is crucial for manipulating and simplifying expressions involving composite functions.
1.
Composition is generally NOT Commutative:
This is perhaps the most important property to remember. In general, $f circ g
eq g circ f$. The order of applying functions matters!
Example:
Let $f(x) = x+1$ and $g(x) = x^2$.
* $f circ g(x) = f(g(x)) = f(x^2) = x^2 + 1$.
* $g circ f(x) = g(f(x)) = g(x+1) = (x+1)^2 = x^2 + 2x + 1$.
Clearly, $x^2 + 1
eq x^2 + 2x + 1$. Hence, $f circ g
eq g circ f$.
There might be specific cases where $f circ g = g circ f$ (e.g., if one function is an identity function, or if they are inverses, or some special functions), but this is an exception, not the rule.
2.
Composition IS Associative:
If we have three functions $f, g, h$, then composition is associative, meaning:
$$ mathbf{(h circ g) circ f = h circ (g circ f)} $$
This means if you want to compose three functions, it doesn't matter which pair you compose first. You can think of it as $h(g(f(x)))$, and the parentheses only dictate the order of evaluation, not the order of application.
3.
Composition with Identity Function:
The identity function, $I(x) = x$, acts like "1" in multiplication.
If $I_A: A o A$ is the identity function on set $A$, and $f: A o B$ is a function, then:
* $f circ I_A = f$
* $I_B circ f = f$ (where $I_B$ is the identity function on set $B$)
This property is useful in understanding inverse functions.
4.
Composition with Inverse Function:
If a function $f: A o B$ has an inverse function $f^{-1}: B o A$, then:
* $f circ f^{-1}(x) = I_B(x) = x$ for all $x in B$.
* $f^{-1} circ f(x) = I_A(x) = x$ for all $x in A$.
This is the very definition of an inverse function.
###
JEE Focus: Finding Domain and Range of Composite Functions
This is where composition gets interesting and challenging for competitive exams. Finding the domain of $g circ f$ requires a two-step process:
1. Identify the domain of the inner function, $D_f$.
2. Find the set of all $x in D_f$ for which $f(x)$ is in the domain of the outer function, $D_g$.
So,
$D_{g circ f} = { x in D_f mid f(x) in D_g }$.
Let's work through some examples to solidify this understanding.
---
### Example 1: Basic Algebraic Functions
Let $f(x) = x^2 - 1$ and $g(x) = frac{1}{x}$.
Part 1: Find $f circ g(x)$ and its domain.
1.
Define $g(x)$ and $f(x)$:
* $f(x) = x^2 - 1$, $D_f = (-infty, infty)$
* $g(x) = frac{1}{x}$, $D_g = (-infty, 0) cup (0, infty)$
2.
Find $f circ g(x)$:
$(f circ g)(x) = f(g(x))$
Substitute $g(x)$ into $f(x)$:
$(f circ g)(x) = fleft(frac{1}{x}
ight) = left(frac{1}{x}
ight)^2 - 1 = frac{1}{x^2} - 1$
3.
Find the domain of $f circ g(x)$:
We need $x in D_g$ AND $g(x) in D_f$.
* Condition 1: $x in D_g implies x
eq 0$.
* Condition 2: $g(x) in D_f$. Since $D_f = (-infty, infty)$, $g(x)$ can be any real number. So, $frac{1}{x}$ can be any real number. This condition imposes no further restrictions on $x$ beyond $x
eq 0$.
Thus, $D_{f circ g} = (-infty, 0) cup (0, infty)$.
Part 2: Find $g circ f(x)$ and its domain.
1.
Find $g circ f(x)$:
$(g circ f)(x) = g(f(x))$
Substitute $f(x)$ into $g(x)$:
$(g circ f)(x) = g(x^2 - 1) = frac{1}{x^2 - 1}$
2.
Find the domain of $g circ f(x)$:
We need $x in D_f$ AND $f(x) in D_g$.
* Condition 1: $x in D_f implies x in (-infty, infty)$ (no restrictions from $f$'s domain itself).
* Condition 2: $f(x) in D_g$. This means $f(x)$ cannot be $0$.
So, $x^2 - 1
eq 0$.
$x^2
eq 1 implies x
eq 1$ and $x
eq -1$.
Combining these, $D_{g circ f} = (-infty, -1) cup (-1, 1) cup (1, infty)$.
---
### Example 2: Functions with Restricted Domains (Typical JEE Problem)
Let $f(x) = sqrt{x-1}$ and $g(x) = x^2+3$.
Part 1: Find $f circ g(x)$ and its domain.
1.
Define $f(x)$ and $g(x)$:
* $f(x) = sqrt{x-1}$. For $f(x)$ to be defined, $x-1 ge 0 implies x ge 1$. So, $D_f = [1, infty)$. The range of $f$ is $R_f = [0, infty)$.
* $g(x) = x^2+3$. $D_g = (-infty, infty)$. The range of $g$ is $R_g = [3, infty)$.
2.
Check if $f circ g$ is defined:
Is $R_g subseteq D_f$?
$R_g = [3, infty)$ and $D_f = [1, infty)$.
Yes, every value in $[3, infty)$ is also in $[1, infty)$. So, $f circ g$ is defined.
3.
Find $f circ g(x)$:
$(f circ g)(x) = f(g(x)) = f(x^2+3)$
Substitute $x^2+3$ into $f(x)$:
$(f circ g)(x) = sqrt{(x^2+3) - 1} = sqrt{x^2+2}$
4.
Find the domain of $f circ g(x)$:
We need $x in D_g$ AND $g(x) in D_f$.
* Condition 1: $x in D_g implies x in (-infty, infty)$.
* Condition 2: $g(x) in D_f$. This means $g(x) ge 1$.
So, $x^2+3 ge 1$.
$x^2 ge -2$. This inequality is true for all real numbers $x$, since $x^2$ is always non-negative.
Combining these, $D_{f circ g} = (-infty, infty)$.
Self-check: The expression $sqrt{x^2+2}$ is also defined for all real $x$ because $x^2+2$ is always positive. This matches our derivation.
Part 2: Find $g circ f(x)$ and its domain.
1.
Check if $g circ f$ is defined:
Is $R_f subseteq D_g$?
$R_f = [0, infty)$ and $D_g = (-infty, infty)$.
Yes, every value in $[0, infty)$ is also in $(-infty, infty)$. So, $g circ f$ is defined.
2.
Find $g circ f(x)$:
$(g circ f)(x) = g(f(x)) = g(sqrt{x-1})$
Substitute $sqrt{x-1}$ into $g(x)$:
$(g circ f)(x) = (sqrt{x-1})^2 + 3 = (x-1) + 3 = x+2$.
3.
Find the domain of $g circ f(x)$:
We need $x in D_f$ AND $f(x) in D_g$.
* Condition 1: $x in D_f implies x ge 1$.
* Condition 2: $f(x) in D_g$. This means $sqrt{x-1}$ must be a real number, which is already handled by $D_f$. $D_g = (-infty, infty)$, so any real output from $f(x)$ is valid for $g$.
Combining these, $D_{g circ f} = [1, infty)$.
Self-check: While the final expression for $g circ f(x)$ is $x+2$, which seems to be defined for all real numbers, its actual domain is restricted by the *original* functions' domains. This is a common trap in JEE! You MUST always consider the domain of the inner function first. The domain of $x+2$ derived from the composite function definition is not necessarily the true domain of $g circ f$.
---
### Example 3: Composition involving Trigonometric and Logarithmic Functions
Let $f(x) = ln(x-2)$ and $g(x) = sin x$.
Part 1: Find $f circ g(x)$ and its domain.
1.
Define $f(x)$ and $g(x)$:
* $f(x) = ln(x-2)$. For $f(x)$ to be defined, $x-2 > 0 implies x > 2$. So, $D_f = (2, infty)$. The range of $f$ is $R_f = (-infty, infty)$.
* $g(x) = sin x$. $D_g = (-infty, infty)$. The range of $g$ is $R_g = [-1, 1]$.
2.
Check if $f circ g$ is defined:
Is $R_g subseteq D_f$?
$R_g = [-1, 1]$ and $D_f = (2, infty)$.
No, $[-1, 1]$ is NOT a subset of $(2, infty)$. The values $sin x$ outputs (between -1 and 1) are not valid inputs for $f(x)$ which requires inputs greater than 2.
Therefore,
$f circ g(x)$ is undefined.
(It's important to state this explicitly in JEE if the composition is not defined for any real $x$).
Part 2: Find $g circ f(x)$ and its domain.
1.
Check if $g circ f$ is defined:
Is $R_f subseteq D_g$?
$R_f = (-infty, infty)$ and $D_g = (-infty, infty)$.
Yes, $(-infty, infty)$ is a subset of $(-infty, infty)$. So, $g circ f$ is defined.
2.
Find $g circ f(x)$:
$(g circ f)(x) = g(f(x)) = g(ln(x-2))$
Substitute $ln(x-2)$ into $g(x)$:
$(g circ f)(x) = sin(ln(x-2))$.
3.
Find the domain of $g circ f(x)$:
We need $x in D_f$ AND $f(x) in D_g$.
* Condition 1: $x in D_f implies x > 2$.
* Condition 2: $f(x) in D_g$. This means $ln(x-2)$ must be a real number. Since $D_g = (-infty, infty)$, $ln(x-2)$ can be any real number, which is true as long as $ln(x-2)$ is defined. The condition for $ln(x-2)$ to be defined is $x-2 > 0 implies x > 2$.
Combining these, $D_{g circ f} = (2, infty)$.
---
### Advanced Application: Decomposing a Function
Sometimes, you might be given a complex function and asked to express it as a composition of simpler functions. This is like reverse engineering the assembly line.
Example: Express $h(x) = sin(sqrt{x^2+1})$ as a composition of three functions.
Let's break down the operations performed on $x$ in sequence:
1. $x$ is squared and 1 is added: $x^2+1$. Let this be $f_1(x) = x^2+1$.
2. The result is square-rooted: $sqrt{x^2+1}$. Let this be $f_2(u) = sqrt{u}$.
3. The result is put into a sine function: $sin(sqrt{x^2+1})$. Let this be $f_3(v) = sin v$.
So, if we define:
* $f_1(x) = x^2+1$
* $f_2(x) = sqrt{x}$
* $f_3(x) = sin x$
Then, $h(x) = f_3(f_2(f_1(x))) = (f_3 circ f_2 circ f_1)(x)$.
### Conclusion
Composition of functions is a core concept that allows us to build complex functions from simpler ones. For JEE, the critical aspect is not just finding the algebraic expression of the composite function, but meticulously determining its
domain and range by considering the conditions for both the inner and outer functions. Always remember to check if the range of the inner function aligns with the domain of the outer function for the composition to be valid. Master these steps, and you'll navigate composition problems with confidence!