Continuity: Learn It 3

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Continuity: Learn It 3

Continuity Over an Interval

Now that we have explored the concept of continuity at a point, let’s extend it to continuity over an interval. A function is continuous over an interval if you can trace it without lifting your pencil between any two points within that interval.

Open Interval: A function is continuous on an open interval $(a,b)$ if it is continuous at every point within that interval.
Closed Interval: A function is continuous on a closed interval $[a,b]$ if it is continuous on $(a,b)$ , continuous from the right at $a$ , and continuous from the left at $b$ .

continuity over an interval

A function is continuous over an open interval if it is continuous at every point in the interval.
A function $f(x)$ is continuous over a closed interval of the form $[a,b]$ if it is continuous at every point in $(a,b)$ and is continuous from the right at $a$ and is continuous from the left at $b$ .
Analogously, a function $f(x)$ is continuous over an interval of the form $(a,b]$ if it is continuous over $(a,b)$ and is continuous from the left at $b$ .

Requiring that $\underset{x\to a^+}{\lim}f(x)=f(a)$ and $\underset{x\to b^-}{\lim}f(x)=f(b)$ ensures that we can trace the graph of the function from the point $(a,f(a))$ to the point $(b,f(b))$ without lifting the pencil. If, for example, $\underset{x\to a^+}{\lim}f(x)\ne f(a)$ , we would need to lift our pencil to jump from $f(a)$ to the graph of the rest of the function over $(a,b]$ .

How To: Determine Continuity Over an Interval

Check Continuity on Open Interval: Verify the function is continuous at all points within $(a, b)$ .
Check Right Continuity at $a$ : Ensure $\lim_{x \to a^+} f(x) = f(a).$
Check Left Continuity at $b$ : Ensure $\lim_{x \to b^-} f(x) = f(b)$ .

State the interval(s) over which the function $f(x)=\dfrac{x-1}{x^2+2x}$ is continuous.

Since $f(x)=\frac{x-1}{x^2+2x}$ is a rational function, it is continuous at every point in its domain.

The domain of $f(x)$ is the set $(−\infty ,-2) \cup (-2,0) \cup (0,+\infty)$ .

Thus, $f(x)$ is continuous over each of the intervals $(−\infty ,-2), \, (-2,0)$ , and $(0,+\infty)$ .

State the interval(s) over which the function $f(x)=\sqrt{4-x^2}$ is continuous.

From the limit laws, we know that $\underset{x\to a}{\lim}\sqrt{4-x^2}=\sqrt{4-a^2}$ for all values of $a$ in $(-2,2)$ .

We also know that $\underset{x\to -2^+}{\lim}\sqrt{4-x^2}=0$ exists and $\underset{x\to 2^-}{\lim}\sqrt{4-x^2}=0$ exists.

Therefore, $f(x)$ is continuous over the interval $[-2,2]$ .