Other Theorems

Darboux’s Theorem

Let $f$ be differentiable on $[a,b]$, $f'(a)\neq f'(b)$ and $u$ is strictly between $f'(a)$ and $f'(b)$:

$$\exists c \in (a,b)\;\text{s.t.}\,f'(c)=u$$

Proof Hint

Use $g(x)=ux - f(x)$ and follow the proof pattern of IVT proof. $$

Rolle’s Theorem

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. And $f(a)=f(b)$. Then:

$$\exists c\in(a,b)\;\text{s.t.}\; f'(c)=0$$

Proof Hint

By Extreme Value Theorem, maximum and minimum exists for $f$. $$

Consider $2$ cases: $$

1. Both minimum and maximum exist at $a$ and $b$.
2. One of minimum or maximum occurs in $(a,b)$.

Mean Value Theorem

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then:

$$\exists c\in(a,b)\;\text{s.t.}\; f'(c)=\cfrac{f(b)-f(a)}{b-a}$$

Proof Hint

• Define $g(x)=f(x)-\Big(\frac{f(a)-f(b)}{a-b}\Big)x$
• $g(a)$ will be equal to $g(b)$
• Use Rolle’s Theorem for $g$

Cauchy’s Mean Value Theorem

Let $f$ and $g$ be continuous on $[a,b]$ and differentiable on $(a,b)$, and $\forall x \in (a,b)\;g'(x) \neq 0$ Then:

$$\exists c\in(a,b)\;\text{s.t.}\; \frac{f'(c)}{g'(c)}=\cfrac{f(b)-f(a)}{g(b)-g(a)}$$

Proof Hint

• Define $h(x)=f(x)-\Big(\frac{f(a)-f(b)}{g(a)-g(b)}\Big)g(x)$
• $h(a)$ will be equal to $h(b)$
• Use Rolle’s Theorem for $h$

Mean value theorem can be obtained from this when $g(x)=x$. $$

L’Hopital’s Rule

Note

Be careful with the pronunciation.

• It’s not “Hospital’s Rule”, there are no “s”
• It’s not “Hopital’s Rule” either, there is a “L’”

Learn the correct pronounciation from this video on YouTube.

L’Hopital’s Rule can be used when all of these conditions are met. (here $\delta$ is some positive number). Select the appropriate $x$ range (as in the limit definition), say $I$. $$

1. Either of these conditions must be satisfied
   • $f(a)=g(a)=0$
   • $\lim{f(x)}=\lim{g(x)}=0$
   • $\lim{f(x)}=\lim{g(x)}=\infty$
2. $f,g$ are continuous on $x\in I$ (closed interval)
3. $f,g$ are differentiable on $x\in I$ (open interval)
4. $g'(x) \neq 0$ on $x\in I$ (open interval)
5. $\lim\limits_{x\to a^{\text{+}}}{\frac{f'(x)}{g'(x)}}=L$

Then: $\lim\limits_{x\to a^{\text{+}}}{\frac{f(x)}{g(x)}}=L$ $$

Here, $L$ can be either a real number or $\pm \infty$. And it is valid for all types of “x limits”.

Proof Hint

L’Hopital’s rule can be proven using Cauchy’s Mean Value Theorem.