Q.Suppose that for a function $f(x)$, $f'(x)\\le1$ for all $1\\le x\\le4$. Show that $f(4)-f(1)\\le3$.

Both theorems below guarantee the existence of a point where the tangent behaves in a special way — they do not by themselves tell you how many such points there are, or give a formula beyond the equation used to solve for them.

Intermediate Value Theorem. If $f$ is continuous on $[a,b]$, then $f$ takes every value between $f(a)$ and $f(b)$ somewhere in $[a,b]$.

Rolle's Theorem. If $f(x)$ is continuous on the closed interval $[a,b]$, differentiable on the open interval $(a,b)$, and $f(a)=f(b)$, then there is at least one $c\in(a,b)$ with $f'(c)=0$.

Geometric meaning: if the endpoint heights match, somewhere in between the tangent must be horizontal (parallel to the $x$-axis).

Rolle's theorem can also be used indirectly, without solving an equation, to bound how many real roots an equation can have in an interval: if $f$ had two roots $\alpha<\beta$ in $(a,b)$ with $f$ continuous/differentiable there, Rolle's theorem would force a zero of $f'$ strictly between them — so if $f'(x)\ne0$ throughout $(a,b)$, $f$ can have at most one root there.

Failure modes for Rolle's theorem (why it does not apply): continuity fails on $[a,b]$ (e.g. an undefined or infinite point inside the interval), differentiability fails somewhere in $(a,b)$ (e.g. a corner, as in $|x|$-type functions), or simply $f(a)\ne f(b)$.

Lagrange's Mean Value Theorem (LMVT). If $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ (with $f(a),f(b)$ not necessarily equal), then there is at least one $c\in(a,b)$ with

Rolle's theorem is the special case $f(a)=f(b)$ (LMVT with right side $=0$) — it is sometimes called the "rotated Rolle's theorem."

Geometric meaning: the tangent at some interior point is parallel to the secant joining the two endpoints — equivalently, the instantaneous rate of change equals the average rate of change over $[a,b]$ at some interior instant.

Consequences of LMVT (used throughout the rest of the chapter). …