*This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.*

# The Composition of Continuous Functions

On the Properties of Continuous Functions page, we looked at some very important theorems regarding combining various functions. We saw that if $f : A \to \mathbb{R}$, and $g : A \to \mathbb{R}$ were both continuous functions at $c \in A$, and $k \in \mathbb{R}$ then:

- $f + g$ is continuous at $c \in A$.

- $f - g$ is continuous at $c \in A$.

- $k f$ is continuous at $c \in A$.

- $fg$ is continuous at $c \in A$.

- $\frac{f}{g}$ is continuous at $c \in A$ provided $g(c) \neq 0$.

- $\mid f \mid$ is continuous at $c \in A$.

- $\sqrt{ f }$ is continuous at $c \in A$ if $f(x) ≥ 0$ for $x$ near $c$.

We will now look at a theorem regarding the continuity of a compositive function, which says that for $A, B \subseteq \mathbb{R}$, and for two functions, $f : A \to \mathbb{R}$ and $g : B \to \mathbb{R}$ such that $f(A) \subseteq B$ where $f$ is continuous at $c \in A$ and $g$ is continuous at $b = f(c) \in B$, then $g \circ f : A \to \mathbb{R}$ is continuous at $c \in A$.

Theorem 1: Let $f : A \to \mathbb{R}$ and $g : B \to \mathbb{R}$. If $f(A) \subseteq B$, and $f$ is continuous at $c \in A$ and $g$ is continuous at $b = f(c) \in B$, then $g \circ f : A \to \mathbb{R}$ is continuous at $c \in A$. |