Linear Functions Continuous at a Single Point Implies Continuous Everywhere
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We have previously discussed the continuity at a point of a function and the different types of discontinuities that exist (see this post for more information: Continuity at a Point). In this tutorial, we will address the continuity on an interval of a function.
What is Continuity on an Interval?
A function f is continuous on an interval if it is continuous at every number in the interval.
The following types of functions are continuous at every number in their domains; in other words, they are continuous on their domains.
- polynomials (continuous everywhere on \mathbb{R} ).
- rational functions (continuous where they are defined).
- root functions (continuous where they are defined).
- trigonometric functions (continuous where they are defined).
- inverse trigonometric functions (continuous where they are defined).
- exponential functions (continuous everywhere on \mathbb{R} ).
- logarithmic functions (continuous on the domain of positive, real numbers).
Examples
Let's now take a look at a few examples illustrating the concept of continuity on an interval.
Example 1: Finding Continuity on an Interval
Find the interval over which the function f(x)= 1- \sqrt{4- x^2} is continuous.
Here is what this function looks like:
We know that this is a root function which is defined on the domain of real numbers provided that :
\sqrt{4- x^2}>= 0 \Longleftrightarrow 4-x^2>= 0 \Longleftrightarrow (2-x)(2+x)>= 0
In addition, we can see that the inequality holds as long as either
2-x>= 0 and 2+x>= 0
or
2-x<= 0 and 2+x<= 0
The first set of inequalities implies that:
x<= 2 and x>= -2
which means that:
x \in [-2, 2]
The second set of inequalities implies that:
x>= 2 and x<= -2
but, there is no solution to this.
In conclusion, the interval of continuity of the function discussed above is x \in [-2, 2] .
Example 2: Finding Continuity on an Interval
Determine the interval on which the function f(x)= \frac{x-3}{x^2+ 2x} is continuous.
Let's take a look at the function above:
First of all, this is a rational function which is continuous at every point in its domain.
Secondly, the domain of this function is x \in \mathbb{R} , provided that
x^2+ 2x \neq 0 \Longleftrightarrow x(x+2) \neq 0 \Longleftrightarrow x \neq 0, x \neq -2
Therefore, the function is continuous on the interval (-\infty, -2), (-2, 0) and (0, \infty) .
Example 3: Finding Continuity on an Interval
Find the interval of continuity for the following function f(x)= \begin{cases} 1-x^2, \text{ if } x<= 1 \\ \ln x, \text{ if } x> 1 \end{cases}
f(x)= 1-x^2 is continuous for all x<= 1 , since it is a polynomial function.
f(x)= \ln x is a logarithmic function which is continuous for all x> 0 . Therefore, f(x)= \ln x is continuous for all x> 1 .
You many also be wondering, what happens when x= 1 ?
Well, that is a great question. We compute the left and right limits and we see that:
\displaystyle\lim_{x \to 1^-}{f(x)}= \displaystyle\lim_{x \to 1^-}{(1-x^2)}= 0 = f(0)
\displaystyle\lim_{x \to 1^+}{f(x)}= \displaystyle\lim_{x \to 1^+}{(\ln x)}= \ln 1= 0 = f(0)
As a result,
\displaystyle\lim_{x \to 1}{f(x)}= 0
Therefore, f(x)= \begin{cases} 1-x^2, \text{ if } x<= 1 \\ \ln x, \text{ if } x> 1 \end{cases} is continuous on the entire set of real numbers, (-\infty, \infty) .
Practice Problems
Now it's your turn to put into practice the ideas explained in this tutorial. Find the interval of continuity of the following functions:
- f(x)= \sqrt{1+ \frac{1}{x}}
- f(x)= 2x^4- 3x^2+ 5x- 2
- Determine the interval on which f(x)= \frac{2x- 5}{x^2+ 3x} is continuous.
- f(x)= 5^x- 3
- f(x)= 5\log_{10}{x}- 12
Practice Solutions
- x \in \mathbb{R} such that x \in [-1, 0) \cup (0, \infty)
- x \in \mathbb{R}
- The answer is x \in (-\infty, -3) \cup (-3, 0) \cup (0, \infty)
- x \in \mathbb{R}
- x \in (0, \infty)
Source: https://mathleverage.com/continuity-on-an-interval/
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