### Functions: Introduction to functions

### Arithmetic operations for continuity

We discuss some methods to create a new continuous function from known continuous functions.

Continuity of sums, products, quotients and compositions of continuous functions

Let #a# be a real number.

- Suppose #f# and #g# are functions continuous in #a#. Then, the functions #f+g# and #f\cdot g# are also continuous in #a#. If #g(a)\ne0#, then we have the same for #\dfrac{f}{g}#.
- Suppose #f# is a function in #a# and that #g# is a function continuous in #f(a)#. Then the composition # {g}\circ{f}# is continuous in #a#.

The first statement follows from the rules of calculation for the sum, the product and the quotient of two limits. For example, for the sum, we have to show that #\lim_{x\to a} (f+g)(x) = (f+g)(a)#. This follows from the following steps, in which we use the definition of the *sum of two functions, *a *rule of calculation for limits*, the *continuity* of #f# and #g# in #a# and once again the definition of the *sum of two* functions #g# in #a#.

\[\begin{array}{rcl}\lim_{x\to a} (f+g)(x) &=& \lim_{x\to a} (f(x)+g(x))\\ &=& \lim_{x\to a} f(x)+ \lim_{x\to a} g(x) \\ &=& f(a)+ g(a)\\ &=& (f+g)(a)\end{array}\]

The second statement we prove using *Limits of continuous functions.* We write #b=f(a)#. Because #f# is continuous in #a#, we have #\lim_{x\to a}f(x) = b#. According to the theory given, we have #\lim_{x\to a}g(f(x)) = g(b)#. But this can be rewritten as #\lim_{x\to a}g\circ f(x) = g\circ f(a)#. This is the definition of continuity of #g\circ f# in #a#.

Continuity of power functions

Let #d# be a real number. If #f# is a function that is continous in #a# with #f(a)>0#, then #f(x)^d# is also continuous in #a#.

This follows from the statement: the function #x^d# is continuous on #\ivoo{0}{\infty}#.

Indeed the original statement follows, by applying rule 2, the *continuity of the composition of continuous functions,* with #g(x)=x^d#.

For a complete proof, we must now prove the statement that #x^d# is continuous. We will only do this only in the case #d# is rational.

When #d# is an integer #n\in\mathbb{Z}\backslash\{0\}#, then the statement follows from applying rule 1: the product of two or more continuous functions is again a continuous function. The rule speaks of the product of two functions, but by applying the rule repeatedly, we see that it also applies to products of more functions. Indeed, we can write #x^n# as a product of continuous functions on #\ivoo{0}{\infty}# : \[n\gt0:\qquad x^n=\underbrace{x\cdot x\cdots x}_{n\text{ terms}}\qquad n\lt0:\qquad x^n=\underbrace{\frac{1}{x}\cdot\frac{1}{x}\cdots\frac{1}{x}}_{n\text{ terms}}\tiny.\]

When #d# has the form #1/m# for an integer #m\in\mathbb{Z}\backslash\{0\}#, then we can write #x^d# as #\sqrt[m]{x}#. From this we know that she is continuous on #\ivoo{0}{\infty}#. If #d# is of the form #n/m#, we can apply rule 1 again. Then it is a product of root functions: \[\qquad x^{n/m}=\underbrace{\sqrt[m]{x}\cdot \sqrt[m]{x}\cdots \sqrt[m]{x}}_{n\text{ terms}}\qquad\tiny,\], hence, also continuous on #\ivoo{0}{\infty}#.

Indeed,

\[

\begin{array}{rcl}

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

&&\phantom{xyzuvw}\color{blue}{\text{composition of functions}} \\

&=& 2^{g(x)} \\

&&\phantom{xyzuvw}\color{blue}{\text{function rule of }f\text{ entered}} \\

&=&2^{x+1} \\

&&\phantom{xyzuvw}\color{blue}{\text{function rule of }g\text{ entered}} \\

\end{array}\]

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