UPSKILL MATH PLUS
Learn Mathematics through our AI based learning portal with the support of our Academic Experts!
Learn moreGiven 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 resulting function.
Example:
If \(f(x) = x -6\) and \(g(x) = x^2\), then find \(f \circ g\).
Solution:
\((f \circ g)(x)\) \(=\) \(f\left(g(x)\right)\)
\(=\) \(f\left(x^2\right)\)
\(=\) \(x^2 -6\)
- A function can also be composed of 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 functions is not commutative. That is, \(f \circ g\) \(\neq\) \(g \circ f\).
Important!
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 \(g\) is not commutative.
Register for free to see more content