UPSKILL MATH PLUS

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### Theory:

Given two functions $$f(x)$$ and $$g(x)$$, then the composition $$f \circ g$$ is computed as follows:
Working rule:
Step 1: Rewrite the given composition as $$(f \circ g)(x)$$ $$=$$ $$f\left(g(x)\right)$$.

Step 2: Using the individual functions as a reference, replace the variable $$x$$ in the outside function with the inside function.

Step 3: Simplify the obtained function.
Example:
If $$f(x) = x -6$$ and $$g(x) = x^2$$, then find $$f \circ g$$.

Solution:

$$(f \circ g)(x)$$ $$=$$ $$f\left(g(x)\right)$$

$$=$$ $$f\left(x^2\right)$$

$$=$$ $$x^2 -6$$
Important!
• A function can also be composed with itself. If $$f$$ is a function, then its composition with itself is given by $$(f \circ f)(x)$$ $$=$$ $$f\left(f(x)\right)$$.
• The composition of function is not commutative. That is, $$f \circ g$$ $$\neq$$ $$g \circ f$$.
Example:
If $$f(x) = x - 6$$ and $$g(x) = x^2$$, then check whether the composition of the two functions are commutative.

Solution:

First, let us find $$(f \circ g)$$.

$$(f \circ g)(x)$$ $$=$$ $$f\left(g(x)\right)$$

$$=$$ $$f\left(x^2\right)$$

$$=$$ $$x^2 -6$$

Now, let us find $$(g \circ f)$$.

$$(g \circ f)(x)$$ $$=$$ $$g\left(f(x)\right)$$

$$=$$ $$g\left(x - 6\right)$$

$$=$$ $$\left(x -6\right)^2$$

$$=$$ $$x^2 - 12x + 36$$

It is observed that $$f \circ g$$ $$\neq$$ $$g \circ f$$.

Therefore, the composition of the two functions $$f$$ and $$g$$ are not commutative.