Theorem 6.4.1. Rolle’s Theorem.
Suppose that \(f:[a,b]\to\mathbb{R}\) is continuous on \([a,b]\) and differentiable on \((a,b)\) and has \(f(a)=f(b)\text{.}\) Then there is some \(c\in (a,b)\) with \(f'(c)=0\)
Section 6.4 The Mean Value Theorem and Company
We start with what sort of looks like it might, if you squinted just right, be considered a converse to Fermat’s Theorem.
Remark 6.4.3.
The hypothesis
\(f:[a,b]\to\mathbb{R}\)\([a,b]\)\((a,b)\)is known as the hypothesis of the Mean Value Theorem. I’ll call it HMVT for short.
Theorem 6.4.4. Cauchy’s Version of the Mean Value Theorem.
If \(f,g:[a,b]\to \mathbb{R}\) each satisfy the HMVT, then there is \(c\in (a,b)\) so that
\begin{equation*}
(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c)\ \ \ .
\end{equation*}
Checkpoint 6.4.5.
If Theorem 6.4.4 seems hard to remember, try writing it as an equation with \(f\)s on one side and \(g\)s on the other.
Checkpoint 6.4.6.
Corollary 6.4.7.
If \(f:[a,b]\to\mathbb{R}\) satisfies HMVT and has \(f'(x)\gt 0\) for all \(x\in (a,b)\text{,}\) then \(f\) is increasing on \((a,b)\) and nondecreasing on \([a,b]\text{.}\)
Corollary 6.4.8.
If \(f:[a,b]\to\mathbb{R}\) satisfies HMVT and has \(f'(x)=0\) for all \(x\in (a,b)\text{,}\) then \(f\) is constant on \([a,b]\text{.}\)
Corollary 6.4.9. Standard Statement of the Mean Value Theorem.
If \(f[a,b]\to \mathbb{R}\) satisfies the HMVT, then there is \(c\in (a,b)\) so that
\begin{equation*}
\frac{f(b)-f(a)}{b-a}=f'(c)\ \ \ .
\end{equation*}
Proof.
See Checkpoint 6.4.5.
We won’t say a whole lot about Taylor’s Theorem, other than to state it as a corollary of the Mean Value Theorem.
Theorem 6.4.10. Taylor’s Theorem.
For any function \(f:[a,b]\to \mathbb{R}\) which has \(n\) derivatives on \((a,b)\) and whose \(n-1\) derivative is continuous on \([a,b]\text{,}\) there is \(c\in (a,b)\) so that
\begin{equation*}
f(b)=f(a)+f'(a)(b-a)+\frac{1}{2}f''(a)(b-a)^2+\cdots+\frac{1}{(n-1)!}f^{(n-1)}(a)(b-a)^{n-1}+\frac{1}{n!}f^{(n)}(c) (b-a)^n\ \ \ .
\end{equation*}
Checkpoint 6.4.11.
The statement of Taylor’s Theorem can be a little hard to parse. Try rewriting the equation by replacing \(b\) with \(x\text{.}\)
