The Intermediate Value Theorem AB & BC

If $f$ is continuous on the closed interval $[a, b]$, and $N$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c$ in $(a, b)$ with $f(c) = N$.

Three parts are load-bearing:

1. Continuity on a closed interval. Not at a point, not on an open interval. A single break anywhere inside can let the function jump the value.
2. $N$ strictly between the endpoint values. Outside that range the theorem says nothing.
3. At least one $c$. Existence only.

The picture is the whole content: an unbroken curve from height $f(a)$ to height $f(b)$ must cross every height in between, because there is no way to get from one to the other without passing through.

The common use is showing an equation has a solution. To show $x^3 + x - 1 = 0$ has a root in $[0, 1]$:

1. $f(x) = x^3+x-1$ is a polynomial, hence continuous on $[0, 1]$. State this.
2. $f(0) = -1$ and $f(1) = 1$.
3. $0$ lies between $-1$ and $1$.
4. By the IVT there is a $c$ in $(0,1)$ with $f(c) = 0$.

Step 1 is where marks are lost. A justification that skips the continuity statement is incomplete even when the arithmetic is right, because the theorem's hypothesis was never checked.

Note that $N = 0$ is a special case, not the definition. The theorem applies to any intermediate value, so it can equally show that $f(c) = 0.4$ for some $c$.

The conclusion is bare existence, and four common overreaches go beyond it:

1. Uniqueness. There may be many such $c$. The IVT never says "exactly one".
2. Location. It does not say $c$ is at the midpoint or anywhere in particular.
3. Extrema. Maxima and minima are the Extreme Value Theorem's business, not the IVT's.
4. Necessity. If the hypotheses fail, the conclusion may still be true. A discontinuous function can hit intermediate values anyway; the theorem simply does not guarantee it.

That last point is worth stating carefully: failing the hypotheses means "no guarantee", not "no root".

Two failures come up repeatedly.

An asymptote inside the interval. $f(x) = \frac{1}{x}$ has $f(-1) = -1$ and $f(1) = 1$, yet takes no value at 0 and never equals 0 anywhere. The IVT does not apply because $f$ is not continuous on $[-1, 1]$; it is not even defined at 0. Applying it anyway "proves" a root that does not exist.

Endpoints not evaluated. The theorem compares $f(a)$ and $f(b)$, so both must be computed. Checking that a function is continuous but never comparing the endpoint values leaves the interval range unknown, and $N$ cannot be confirmed to lie between them.

Both failures are caught by the same habit: write the hypotheses out and tick each one before invoking the conclusion.