The mean value theorem is rarely used to find $c$. Its power is in inequalities: since we do not know where $c$ is, we bound $f'(c)$ over the whole interval, and that bound controls how far $f$ can move. This is the cleanest route to uniform continuity.
🔑 Key idea
If $|f'(x)| \le K$ for all $x$ in an interval $I$, then $|f(x) - f(y)| \le K|x - y|$ for all $x, y \in I$ ($f$ is Lipschitz with constant $K$). Then $\delta = \frac{\varepsilon}{K}$ works for every point at once, so $f$ is uniformly continuous on $I$. The best constant is $K = \sup_I |f'|$.
💡 Why it works
For $x \ne y$ in $I$, the MVT gives $f(x) - f(y) = f'(c)(x - y)$ for some $c$ between them, and $c \in I$. So $|f(x) - f(y)| = |f'(c)||x - y| \le K|x - y|$. Then $|x - y| < \frac{\varepsilon}{K}$ gives $|f(x) - f(y)| < \varepsilon$, and $\delta = \frac{\varepsilon}{K}$ does not depend on the points. $\blacksquare$ Taking $f = \sin$ and $K = 1$ gives the useful $|\sin x - \sin y| \le |x - y|$ for all real $x, y$.
$f(x) = \sqrt{x}$ on $[4, 9]$: the slope is largest at the left end, $\frac{1}{4}$, so the graph never rises faster than a line of slope $\frac{1}{4}$.
The same argument proves the facts about monotonicity that calculus takes for granted. If $f' > 0$ on an interval, then for $x < y$ the MVT gives $f(y) - f(x) = f'(c)(y - x) > 0$: the function increases. If $f' = 0$ everywhere on an interval, then $f(y) - f(x) = 0$ for all $x, y$: the function is constant. Notice that each conclusion needs information about $f'$ at every point between $x$ and $y$, not at one point.
Worked examples
✏️ Worked example: The best Lipschitz constant
Find the smallest $K$ with $|x^{2} - y^{2}| \le K|x - y|$ for all $x, y \in [-3, 2]$.
$|f'(x)| = 2|x|$, largest at the end farther from $0$, $x = -3$.
Answer: $K = 6$.
✏️ Worked example: A δ that works everywhere
On $[1, 4]$, $\frac{1}{x}$ has $|f'(x)| = \frac{1}{x^{2}} \le 1$. What δ does the bound give for $\varepsilon = 0.05$?
$|f(x) - f(y)| \le 1 \cdot |x - y| < \delta$.
Answer: $\delta = 0.05$, for every pair of points in $[1, 4]$.
✏️ Worked example: Estimating a square root
Use the MVT to bound $\sqrt{102} - 10$ from above.
$\sqrt{102} - \sqrt{100} = \frac{1}{2\sqrt{c}} \cdot 2$ for some $c \in (100, 102)$.
Since $\sqrt{c} > 10$, this is less than $\frac{2}{20} = \frac{1}{10}$.
Answer: $\sqrt{102} < 10.1$ (in fact $\sqrt{102} \approx 10.0995$).
⚠️ Common mistake
Using $f'$ at one point instead of a bound over the interval. "$f'(4) = \frac{1}{4}$, so $|\sqrt{x} - \sqrt{y}| \le \frac{1}{4}|x - y|$" is right on $[4, 9]$ only because $\frac{1}{4}$ happens to be the largest value of $f'$ there; on $[1, 9]$ it is false (the slope reaches $\frac{1}{2}$ at $x = 1$). The MVT's $c$ could be anywhere between the points, so you must bound $|f'|$ everywhere on the interval.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.