Modules › Continuity and the Intermediate Value Theorem
Continuity and the Intermediate Value Theorem
Continuity at a point and on an interval, types of discontinuity, continuous functions, the IVT.
Lectures 2–3 · Tutorial 3 (first half) · Functions, limits and continuity
Textbook sections: Thomas ch. 1 Functions (1.1–1.3) and ch. 2 Limits and Continuity (2.1–2.6) · Stewart ch. 1 (1.1–1.3) and ch. 2 (2.1–2.6)
Textbook sections: Thomas ch. 1 Functions (1.1–1.3) and ch. 2 Limits and Continuity (2.1–2.6) · Stewart ch. 1 (1.1–1.3) and ch. 2 (2.1–2.6)
Continuity and the Intermediate Value Theorem
Calculus I · Module 3
After this module you can
- Check the three conditions for continuity and find parameters that make a piecewise function continuous.
- Classify discontinuities (removable, jump, infinite).
- Use the IVT to show that an equation has a solution and to count roots.
Continuity
- $f$ is continuous at $a$ if: (1) $f(a)$ is defined, (2) $\lim_{x\to a}f(x)$ exists, (3) $\lim_{x\to a}f(x)=f(a)$.
- Polynomials, rational functions (on their domain), roots, $\sin,\cos,e^x,\ln x$ are continuous; sums, products, quotients and compositions of continuous functions are continuous.
- Piecewise function: continuity at the junction $\iff$ left limit = right limit = value.
- Example: $x^2-1$ ($x<3$), $2ax$ ($x\ge3$) is continuous iff $8=6a$, i.e. $a=\frac43$.
Types of discontinuity
| Type | What happens | Example at $x=0$ |
|---|---|---|
| removable | limit exists, value missing or different | $\frac{x^2-x}{x}$ |
| jump | one-sided limits exist but differ | $\frac{|x|}{x}$, $\lfloor x\rfloor$ at integers |
| infinite | a one-sided limit is $\pm\infty$ | $\frac{1}{x^2}$ |
| oscillating | no one-sided limit | $\sin\frac1x$ |
Intermediate Value Theorem
- If $f$ is continuous on $[a,b]$ and $N$ is between $f(a)$ and $f(b)$, then $f(c)=N$ for some $c\in(a,b)$.
- Root finding: if $f(a)f(b)<0$ there is a root in $(a,b)$.
- Example: $x^3-15x+1$ has $f(-4)=-3$, $f(-1)=15$, $f(1)=-13$, $f(4)=5$: three sign changes, three roots in $[-4,4]$.
- Continuity on the whole closed interval is essential (a jump can skip the value).
Applications
- $e^x=-x$ has a solution: $F(x)=e^x+x$ has $F(-1)<0<F(0)$.
- $(1-x)^3=\sin x$ has a root in $(0,1)$: $F(0)=1>0$, $F(1)=-\sin1<0$.
- The monk and the mountain: let $h(t)=$ (position going up) $-$ (position going down); $h(\text{start})<0<h(\text{end})$ so $h(t)=0$ at some time.
- Bisection: halve the interval keeping the sign change (see the Calculus lab).
Typical mistakes
No sign change$f(a)f(b)>0$ does not prove there is no root (two roots may hide inside).
Forgetting closed intervalIVT needs continuity on all of $[a,b]$.
Product of continuous$fg$ continuous at $0$ does NOT force $f$ and $g$ to be continuous.
Practice next
- Chapter quizzes Ch. 1 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Newton's method and bisection
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Continuity (Computation problems 4, Q2)
Let $g(x)=x+b$ for $x<0$ and $g(x)=\cos x$ for $x\ge0$. Which statement is true?
Show the answer and the solution
Answer: $g$ is continuous at $0$ iff $b=1$, but $g$ is not differentiable at $0$ for any $b$
Continuity: $\lim_{x\to0^-}(x+b)=b$ must equal $\cos0=1$, so $b=1$. One-sided derivatives: left $=1$, right $=-\sin0=0$. They differ, so $g'(0)$ never exists.
Worked example 2 · Intermediate Value Theorem (Tutorial 3, Q5)
For $f(x)=x^3-15x+1$, on which interval do the hypotheses of the IVT guarantee a root (a sign change of the continuous function $f$)?
Show the answer and the solution
Answer: $[3,4]$
$f$ is continuous, so a root is guaranteed on $[a,b]$ when $f(a)f(b)<0$. Here $f(3)=-17$, $f(4)=5$ have opposite signs; for the other intervals the endpoint values have the same sign.
Worked example 3 · Continuity (Tutorial 2, Q8)
For which $a$ is $f(x)=x^2-1$ ($x<3$), $f(x)=2ax$ ($x\ge3$) continuous at every $x$?
Show the answer and the solution
Answer: $\frac{4}{3}$
Continuity at $3$: $\lim_{x\to3^-}(x^2-1)=8$ must equal $f(3)=6a$, so $a=\frac43$.
Worked example 4 · Intermediate Value Theorem (Tutorial 3, Q5)
For $f(x)=x^3-15x+1$, on which interval do the hypotheses of the IVT guarantee a root (a sign change of the continuous function $f$)?
Show the answer and the solution
Answer: $[-4,-3]$
$f$ is continuous, so a root is guaranteed on $[a,b]$ when $f(a)f(b)<0$. Here $f(-4)=-3$, $f(-3)=19$ have opposite signs; for the other intervals the endpoint values have the same sign.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Newton's method and bisection