Continuous function
Construction of continuous functions

Checking the continuity of a given function can be simplified by checking one of the above defining properties for the building blocks of the given function. It is straightforward to show that the sum of two functions, continuous on some domain, is also continuous on this domain. Given f, g: D → R, {\displaystyle f,g\colon D\to \mathbb {R},} then the sum of continuous functions s = f + g {\displaystyle s=f+g} (defined by s (x) = f (x) + g (x) {\displaystyle s(x)=f(x)+g(x)} for all x ∈ D {\displaystyle x\in D}) is continuous in D. {\displaystyle D.}

The same holds for the product of continuous functions, p = f ⋅ g {\displaystyle p=f\cdot g} (defined by p (x) = f (x) ⋅ g (x) {\displaystyle p(x)=f(x)\cdot g(x)} for all x ∈ D {\displaystyle x\in D})is continuous in D. {\displaystyle D.}

Combining the above preservations of continuity and the continuity of constant functions and of the identity function I (x) = x {\displaystyle I(x)=x} on R {\displaystyle \mathbb {R} }, one arrives at the continuity of all polynomial functions on R {\displaystyle \mathbb {R} }, such as f (x) = x 3 + x 2 − 5 x + 3 {\displaystyle f(x)=x^{3}+x^{2}-5x+3} (pictured on the right).

In the same way, it can be shown that the reciprocal of a continuous function r = 1 / f {\displaystyle r=1/f} (defined by r (x) = 1 / f (x) {\displaystyle r(x)=1/f(x)} for all x ∈ D {\displaystyle x\in D} such that f (x) ≠ 0 {\displaystyle f(x)\neq 0})is continuous in D ∖ { x: f (x) = 0 }. {\displaystyle D\setminus \{x:f(x)=0\}.}

This implies that, excluding the roots of g, {\displaystyle g,} the quotient of continuous functions q = f / g {\displaystyle q=f/g} (defined by q (x) = f (x) / g (x) {\displaystyle q(x)=f(x)/g(x)} for all x ∈ D {\displaystyle x\in D}, such that g (x) ≠ 0 {\displaystyle g(x)\neq 0})is also continuous on D ∖ { x: g (x) = 0 } {\displaystyle D\setminus \{x:g(x)=0\}}.

For example, the function (pictured) y (x) = 2 x − 1 x + 2 {\displaystyle y(x)={\frac {2x-1}{x+2}}} is defined for all real numbers x ≠ − 2 {\displaystyle x\neq -2} and is continuous at every such point. Thus, it is a continuous function. The question of continuity at x = − 2 {\displaystyle x=-2} does not arise since x = − 2 {\displaystyle x=-2} is not in the domain of y. {\displaystyle y.} There is no continuous function F: R → R {\displaystyle F:\mathbb {R} \to \mathbb {R} } that agrees with y (x) {\displaystyle y(x)} for all x ≠ − 2. {\displaystyle x\neq -2.}

Since the function sine is continuous on all reals, the sinc function G (x) = sin (x) / x, {\displaystyle G(x)=\sin(x)/x,} is defined and continuous for all real x ≠ 0. {\displaystyle x\neq 0.} However, unlike the previous example, G can be extended to a continuous function on all real numbers, by defining the value G (0) {\displaystyle G(0)} to be 1, which is the limit of G (x), {\displaystyle G(x),} when x approaches 0, i.e., G (0) = lim x → 0 sin x x = 1. {\displaystyle G(0)=\lim _{x\to 0}{\frac {\sin x}{x}}=1.}

Thus, by setting

the sinc-function becomes a continuous function on all real numbers. The term removable singularity is used in such cases when (re)defining values of a function to coincide with the appropriate limits make a function continuous at specific points.

A more involved construction of continuous functions is the function composition. Given two continuous functions g: D g ⊆ R → R g ⊆ R and f: D f ⊆ R → R f ⊆ D g, {\displaystyle g:D_{g}\subseteq \mathbb {R} \to R_{g}\subseteq \mathbb {R} \quad {\text{ and }}\quad f:D_{f}\subseteq \mathbb {R} \to R_{f}\subseteq D_{g},} their composition, denoted as c = g ∘ f: D f → R, {\displaystyle c=g\circ f:D_{f}\to \mathbb {R},} and defined by c (x) = g (f (x)), {\displaystyle c(x)=g(f(x)),} is continuous.

This construction allows stating, for example, that e sin (ln x) {\displaystyle e^{\sin(\ln x)}} is continuous for all x > 0. {\displaystyle x>0.}
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