Continuity and the Intermediate Value Theorem
Without lifting the pencil
A continuous function is often described as one you can draw without lifting your pencil. That is a good picture, but what exactly has to be true at a point for the pencil to stay down?
Below, each function is checked against the three conditions in the definition. Find out which condition fails at each kind of break, then use the last function to repair a break yourself.
For the last function, drag the slider for k until the two pieces meet.
The definition
A function f is continuous at a when
That one equation packs in three conditions: f(a) is defined, the limit exists, and the two are equal. Each kind of discontinuity breaks a different one.
- Removable
The limit exists, but f(a) is missing or has the wrong value. Redefining one point fixes it.
- Jump
The one-sided limits exist but differ, so the limit does not exist. No single value at a can fix it.
- Infinite
At least one side grows without bound, usually at a vertical asymptote.
Polynomials, rational functions, roots, and the trigonometric, exponential and logarithmic functions are continuous at every point of their domains. Sums, products, quotients and compositions of continuous functions are continuous too, where they are defined. That is why direct substitution works so often.
The Intermediate Value Theorem
Suppose f is continuous on [a,b] and N is any number between f(a) and f(b). Then there is at least one c in (a,b) with f(c)=N.
The theorem says a root exists without telling you where. Bisection turns it into a method: check the sign at the midpoint and keep the half where the sign changes. Each step halves the interval that must contain the root. How many steps would it take to pin a root down to three decimal places?
Drag the ochre endpoints along the x-axis, or bisect. The indigo line is the target height N.
| Step | a | b | Midpoint m | f(m) − N | Width |
|---|
Solve it step by step
Check continuity one condition at a time, and decide what the Intermediate Value Theorem guarantees. The score counts steps you answer correctly on the first try.
Common mistakes
- Checking only that f(a) exists
A function can be defined at a and still be discontinuous there, as with a jump or a misplaced point.
- Using the IVT without continuity
\tfrac1x is negative at -1 and positive at 1, yet it is never 0. The theorem needs continuity on the whole closed interval. Choose it in the bisection graph and watch where bisection ends up.
- Reading the IVT backwards
If f(a) and f(b) have the same sign, the theorem guarantees nothing. There may still be roots in between, an even number of them.