Lesson 1.9 · Unit 1 · CED topic 1.12
Confirming continuity over an interval
Continuity at a single point is a local property. Most of the time we care about continuity on a whole interval, because the big theorems of calculus (the Intermediate Value Theorem, the Extreme Value Theorem, and eventually the Fundamental Theorem of Calculus) all begin with the hypothesis "\(f\) is continuous on \([a,b]\)." This lesson shows how to confirm that hypothesis quickly, using the shape of a function's formula instead of testing every point.
Learning objective.
- LIM-2.B. Determine intervals over which a function is continuous.
Essential knowledge.
- LIM-2.B.1. A function is continuous on an interval if the function is continuous at each point in the interval.
- LIM-2.B.2. Polynomial, rational, power, exponential, logarithmic, and trigonometric functions are continuous on all points in their domains.
Continuity on an interval
A function is continuous on an interval if it is continuous at each point in the interval. For an open interval \((a,b)\), that means continuity at every \(x\) with \(a\lt x\lt b\). For a closed interval \([a,b]\), we need continuity at each interior point and one-sided continuity at the endpoints:
\[\lim_{x\to a^+}f(x)=f(a)\qquad\text{and}\qquad\lim_{x\to b^-}f(x)=f(b).\]
At an endpoint only one side of the function exists inside the interval, so only one side can be checked. The graph of a function that is continuous on \([a,b]\) is one unbroken curve from the point \((a,f(a))\) to the point \((b,f(b))\).
Continuity of the familiar functions
You do not have to verify the three conditions point by point for every function. These families are continuous at every point of their domains:
- polynomial functions (continuous on all real numbers);
- rational functions (continuous wherever the denominator is not zero);
- power and root functions, \(x^r\);
- exponential and logarithmic functions (\(e^x\) everywhere, \(\ln x\) for \(x\gt0\));
- trigonometric functions (\(\sin x\) and \(\cos x\) everywhere; \(\tan x\), \(\sec x\) wherever \(\cos x\ne0\); \(\cot x\), \(\csc x\) wherever \(\sin x\ne0\)).
Sums, differences, products, and composites of continuous functions are continuous where they are defined, and a quotient of continuous functions is continuous wherever the denominator is not zero. These properties follow directly from the limit properties of Lesson 1.5. The practical message: a function built from the familiar pieces is continuous on every interval inside its domain. To find where it might fail, find where it is not defined, and check any places where the formula changes.
Find the intervals on which \(f(x)=\dfrac{\sqrt{x-1}}{x^2-9}\) is continuous.
The square root needs \(x-1\ge0\), so \(x\ge1\). The denominator is zero when \(x^2=9\), that is \(x=3\) or \(x=-3\); only \(x=3\) lies in \(x\ge1\). The domain is \([1,3)\cup(3,\infty)\).
Since \(f\) is built from continuous pieces, it is continuous on its domain: on \([1,3)\) and on \((3,\infty)\). (At \(x=1\) the function is continuous from the right, which is all that is possible there.) At \(x=3\) there is an infinite discontinuity.
On what intervals is \(g(x)=\ln(x^2-4)\) continuous?
\(\ln\) requires a positive input: \(x^2-4\gt0\), so \(x\lt-2\) or \(x\gt2\). Thus \(g\) is continuous on \((-\infty,-2)\) and on \((2,\infty)\). The function is not defined (so not continuous) at \(x=\pm2\) or anywhere between.
Piecewise functions
For a piecewise function, each piece is continuous on its own interval because it is built from continuous parts. The only places to check are the boundaries where the formula changes, using the three-condition test of Lesson 1.11.
Is \(f(x)=\begin{cases}x^2+1,&x\lt1\\ 3x-1,&x\ge1\end{cases}\) continuous on \((-\infty,\infty)\)?
Each piece is a polynomial, so \(f\) is continuous for \(x\lt1\) and for \(x\gt1\). At the boundary \(x=1\): \(f(1)=3(1)-1=2\); the left limit is \(1^2+1=2\); the right limit is \(3(1)-1=2\). All three conditions hold, so \(f\) is continuous at \(1\), and therefore continuous on all real numbers.
Why this matters
When a theorem requires "\(f\) is continuous on \([a,b]\)," you verify it by pointing to the type of function (or to a boundary check) and say so: "\(f\) is a rational function whose denominator is nonzero on \([a,b]\), so it is continuous there." Watch for a closed interval that contains a point outside the domain. For example, \(\tan x\) is not continuous on \([0,\pi]\) because it is undefined at \(x=\dfrac\pi2\), so the Intermediate Value Theorem cannot be applied there.
- Continuous on a closed interval \([a,b]\) means continuous inside \((a,b)\), right-continuous at \(a\), and left-continuous at \(b\).
- Polynomial, rational, power, exponential, logarithmic, and trigonometric functions are continuous at every point of their domains.
- Find intervals of continuity by finding the domain (denominators not zero, even roots of nonnegative numbers, logarithms of positive numbers).
- For piecewise functions, check each boundary with the three-condition test.
- Justify continuity by naming the function type and the domain, not by saying "the graph has no breaks."
- On what intervals is \(f(x)=\dfrac{x+2}{x^2-x-6}\) continuous?
Show answer
\(x^2-x-6=(x-3)(x+2)\) is zero at \(x=3\) and \(x=-2\). The function is continuous on \((-\infty,-2)\), \((-2,3)\), and \((3,\infty)\).
- Is \(g(x)=\sqrt{4-x^2}\) continuous on \([-2,2]\)? Explain what happens at the endpoints.
Show answer
Yes. The domain is \(-2\le x\le2\), and \(g\) is continuous at every point of its domain. At \(x=-2\) it is continuous from the right and at \(x=2\) from the left, which is what continuity on a closed interval requires.
- Is \(f(x)=\begin{cases}x^2+1,&x\lt1\\ 3x-1,&x\ge1\end{cases}\) continuous at \(x=1\)? Is it continuous everywhere?
Show answer
\(f(1)=2\), and both one-sided limits equal 2, so \(f\) is continuous at 1. Each piece is a polynomial, so it is continuous elsewhere; \(f\) is continuous on all real numbers.
- Can the Intermediate Value Theorem be applied to \(f(x)=\tan x\) on \([0,\pi]\)? Why or why not?
Show answer
No. \(\tan x\) is undefined at \(x=\dfrac\pi2\) in the interval, so \(f\) is not continuous on \([0,\pi]\), and the hypothesis of the theorem is not satisfied.
- Where is \(h(x)=\sin\dfrac1x\) continuous?
Show answer
\(h\) is a composite of continuous functions wherever \(\dfrac1x\) is defined, so it is continuous on \((-\infty,0)\) and \((0,\infty)\). It is not defined at \(0\), and (as Lesson 1.3 showed) it oscillates there, so it cannot be repaired.
- Find the value of \(k\) that makes \(f(x)=\begin{cases}kx+1,&x\le3\\ x^2-2,&x\gt3\end{cases}\) continuous at \(x=3\). (This is the question type of Lesson 1.13; try it now.)
Show answer
Left side: \(f(3)=3k+1\). Right limit: \(3^2-2=7\). Set \(3k+1=7\), so \(k=2\).