Continuity and guaranteed crossings

Try

Suppose a continuous temperature model starts at 18 degrees and ends at 24 degrees during one hour. Must it equal 21 at some time in between? Must it equal 21 exactly once? The first conclusion follows from continuity; the second needs additional information.

Predict

Predict which problems can be repaired by assigning one missing value: a hole where both branches approach height 4, or a jump with left height 4 and right height 7. Explain why assigning a value cannot change either one-sided limit.

Experiment

For x≠2 let f(x)=(x²−4)/(x−2), and choose f(2)=k. Nearby, f(x)=x+2, so its limit is 4. Compare the three tests below for k=0, k=4 and an unassigned value.

Continuity test at 2	k=0	k=4	Unassigned
Function value exists	yes	yes	no
Finite limit exists	yes	yes	yes
Value equals limit	no	yes	no

Sketch each graph using the same axes. Only the filled point changes. A continuous repair requires k=4, rather than any convenient number.

Understand

A function is continuous at a when f(a) exists, the limit at a exists, and that limit equals f(a). Continuity over a closed interval means continuity throughout its interior with the appropriate one-sided continuity at the endpoints. Polynomials are continuous everywhere; rational functions are continuous where their denominators are nonzero.

The intermediate value theorem, or IVT, states that a continuous function on a closed interval [a,b] takes every value between f(a) and f(b). If a target lies strictly between the endpoint values, it occurs at an interior input. This is an existence guarantee; it supplies neither a location formula nor a count of how often the target occurs.

For p(x)=x³+x−1 on [0,1], p(0)=−1 and p(1)=1. Polynomial continuity and the sign change guarantee a zero strictly between 0 and 1. Directly substituting 1/2 gives 1/8+1/2−1=−3/8, so the root is not simply the interval midpoint. A separate argument about increase could establish uniqueness, but IVT alone does not.

Now test q(x)=1/x on [−1,1]. Its endpoint values have opposite signs, yet q is undefined at zero and never equals zero. The missing continuity hypothesis invalidates the crossing argument. An apparent graph connection across a vertical asymptote is not a root.

For a piecewise formula, match the left limit to the right limit and actual value. If one branch is mx+1 to the left of 2 and the other is x²−1 at and to the right of 2, the equation 2m+1=3 yields m=1. You must check the branch that includes the endpoint, not merely compare arbitrary nearby samples.

Apply

Write IVT solutions in three sentences: identify the continuous function and interval; compute endpoint values and locate the target between them; state the existence conclusion with its limitations. Practice repairing continuity and identifying invalid theorem applications as well as obtaining numerical parameters.

Continue

The next lesson separates very large inputs from outputs that grow without bound near a finite input. Neither type of behavior makes infinity an ordinary function value.