2.0 Algebra of continuous functions

  Continuity and Differentiability

Let us assume two real functions $f(x)$ and $g(x)$ which are continuous at $x=c$. Consider another function $h(x)$ which can be written in following forms:

$(a)$ $h(x)=f(x)+g(x)$

Statement: If $f(x)$ and $g(x)$ are continuous functions at $x=c$, then $h(x)$ is also continuous function at $x=c$.

Proof: We can write $$\begin{equation} \begin{aligned} \mathop {\lim }\limits_{x \to c} h(x) = \;\mathop {\lim }\limits_{x \to c} \left[ {f(x) + g(x)} \right] \\ \mathop {\lim }\limits_{x \to c} h(x) = \;\mathop {\lim }\limits_{x \to c} \left[ {f(x)} \right] + \mathop {\lim }\limits_{x \to c} \left[ {g(x)} \right] \\\end{aligned} \end{equation} $$Since both $f(x)$ and $g(x)$ are continuous at $x=c$, $$\mathop {\lim }\limits_{x \to c} h(x) = \;f(c) + g(c) = (f + g)(c)$$Hence, we can say that $h(x)$ is continuous function at $x=c$.

Using the same concept, we can say that

$(b)$ $f(x)-g(x)$ is continuous at $x=c$.

$(c)$ $f(x).g(x)$ is continuous at $x=c$.

$(d)$ $f(x)/g(x)$ is continuous at $x=c$ provided $g(c) \ne 0$.

$(e)$ $f(g(x))$ is continuous at $x=c$ if both $g(x)$ and $f(x)$ are continuous at $x=c$.

Question 2. Show that the function $f$ defined by $$f(x) = \left| {1 - x + \left| x \right|} \right|$$where $x$ is any real number, is a continuous function.

Solution: We can write this given function $f(x)$ using two functions $g(x)$ and $h(x)$ as shown.$$g(x) = 1 - x + \left| x \right|\;and\;h(x) = \left| x \right|\quad \in real\;values\;of\;x$$Then,$$\begin{equation} \begin{aligned} \left( {hog} \right)(x) = h(g(x)) = h\left( {1 - x + \left| x \right|} \right) \\ h(g(x)) = \left| {1 - x + \left| x \right|} \right| = f(x) \\\end{aligned} \end{equation} $$

As we know that $h(x)$ being the modulus function is a continuous function and $g(x)$ being a sum of a polynomial function and the modulus function is continuous.

Hence, $f(x)$ being a composite of two continuous functions is also a continuous function.