Confirming Continuity over an Interval
Introduced| English |
|---|
| interval/ˈɪntəvl/ |
| domains/dəˈmeɪnz/ |
From one point to a whole stretch
- Continuity at a single point is nice; usually we want it across an interval 区间.
- A function is continuous on an interval if it is continuous at every point inside it.
- That's what lets you draw the whole stretch without lifting your pen.
- Two subtleties: how to treat the endpoints, and which familiar functions are automatically continuous.
Continuous across the whole view
y = ax³ + bx
A polynomial is continuous on every interval — no holes, jumps, or blow-ups anywhere you look.
Open vs. closed intervals
- On an open interval $(a,b)$: check the three-condition continuity at every interior point.
- On a closed interval $[a,b]$: also handle the endpoints, where you can only approach from one side.
- At the left end $a$, require right-continuity: $\displaystyle\lim_{x\to a^+}f(x)=f(a)$.
- At the right end $b$, require left-continuity: $\displaystyle\lim_{x\to b^-}f(x)=f(b)$.
At the left endpoint $a$ of a closed interval $[a,b]$, continuity requires...
You can only approach the left endpoint from the right, so right-continuity is required.
For $f(x)=\sqrt x$ on $[0,9]$, the left-continuity check at the right endpoint needs $\lim_{x\to9^-}\sqrt x$. What is that value?
$\sqrt 9 = 3$, which equals $f(9)$, so $f$ is left-continuous there.
The functions that come "pre-continuous"
- On their natural domains 定义域, these families are continuous everywhere:
- Polynomials — continuous for all real $x$.
- Rational, exponential, logarithmic, and trigonometric functions — continuous wherever they are defined.
- So a rational function is continuous except where its denominator is zero; $\ln x$ is continuous for $x>0$.
Which function is continuous for all real $x$?
Polynomials are continuous everywhere; the others have domain gaps.
Select all function families that are continuous on their entire natural domain.
All four are continuous throughout their domains — the domain just excludes trouble points (e.g. rational at zeros of the denominator).
Why we care: it unlocks theorems
- Continuity on a closed interval is the entry ticket to the big theorems ahead.
- The Intermediate Value Theorem (1.16) and the Extreme Value Theorem (Unit 5) both require continuity on $[a,b]$.
- Skip the continuity check and those theorems simply do not apply — a common exam trap.
- So confirming continuity over an interval is not busywork; it is the hypothesis everything else stands on.
$f(x)=\dfrac1x$ is continuous on the interval $[-1,1]$.
$0\in[-1,1]$ and $f$ blows up there, so it is not continuous on that interval.
Continuity on a closed interval is a required hypothesis of the ____ Value Theorem.
The IVT (and the Extreme Value Theorem) require continuity on $[a,b]$.
"Continuous on its domain" is not the same as "continuous everywhere." $f(x)=\dfrac1x$ is continuous on its domain (all $x\neq0$), yet it is not continuous on $[-1,1]$ because $0$ is in that interval and $f$ blows up there. Always check that the interval avoids the trouble spots.
Is $f(x)=\sqrt{x}$ continuous on $[0,4]$?
- Interior $(0,4)$: $\sqrt x$ is continuous (a standard function on its domain). ✓
- Left endpoint $0$: $\displaystyle\lim_{x\to0^+}\sqrt x=0=f(0)$ — right-continuous. ✓
- Right endpoint $4$: $\displaystyle\lim_{x\to4^-}\sqrt x=2=f(4)$ — left-continuous. ✓
- So $f$ is continuous on the closed interval $[0,4]$.
$f$ is continuous on an interval when it is continuous at every point inside, using one-sided continuity at the endpoints of a closed interval. Polynomials, rational, exponential, logarithmic, and trigonometric functions are continuous on their domains. This interval-continuity is the hypothesis the IVT and EVT depend on.