Understanding Cauchy Mean Value Theorem

Question

I got a hard time to understand the Theorem from Cauchy Mean Value Theorem. Could someone please to help me explain this?

Let I be an open interval and $n$ be a natural number and suppose that the function $f: I \rightarrow R$ has n derivatives. Suppose also that the point $x_0$ in $I$:

$f^{(k)}(x_0) = 0$ for $0 \le k \le n-1$

Then, for each point $x \not = x_0$ in $I$, there is a point $z$ strictly between $x$ and $x_0$ at which

$f(x) = \frac{f^{(n)}(z)}{n!}(x-x_0)^n$

Then my homework is giving $f(x) = (x-2)^5$ with $f(x_0) = f'(x_0) = ... = f^{n-1}(x_0)$ with $x_0 = 2$ and $n=3$, find all possible $z$ to satisfy the Theorem above.

Answer 1

The function $f$ satisfies the assumptions for Taylor's theorem, so $f(x) = P_{n-1}(x) + R_{n-1}(x)$ where $$ P_{n-1}(x) = \sum_{k=0}^{n-1}\frac{f^{(k)(x_0)}}{k!}(x-x_0)^k $$ and $$ R_{n-1}(x) = \frac{f^{(n)}(z)}{n!}(x-x_0)^n $$ for some $z$ between $x$ and $x_0$.

In the case where $f(x)=(x-2)^5$, $x_0=2$, and $n=3$, we have $$ \frac{f^{(n)}(z)}{n!}(x-x_0)^n = \frac{60(z-2)^2}{3!}(x-2)^3 = 10 (z-2)^2(x-2)^3. $$ Equating the above with $f(x)$, we have $$ (x-2)^5 = 10 (z-2)^2(x-2)^3, $$ and hence $$ z = \pm \left(2 +\frac1{\sqrt{10}}(x-2)\right). $$

Answer 2

Observe that

$$f(x)-f(x_0)=\int_{x_0}^x f'(t)dt=(t-x)f'(t)\big|_{x_0}^x-\int_{x_0}^x(t-x)f''(t)dt\\=(t-x)f'(t)\big|_{x_0}^x-\frac{(t-x)^2}2f''(t)\big|_{x_0}^x+\int_{x_0}^x\frac{(t-x)^2}2f'''(t)dt=\ldots\\=\sum_{k=1}^{n-1}(-1)^{k-1}\frac{(t-x)^k}{k!}f^{(k)}(t)|_{x_0}^x+(-1)^{n-1}\int_{x_0}^x\frac{(t-x)^{n-1}}{(n-1)!}f^{(n)}(t) dt$$

where I used integration by parts each time, choosing $-x$ as the constant of integration of $1$. Then when $f^{(k)}(x_0)=0$ for all $k\in\{0,1,\ldots,n-1\}$ the above reduces to

$$f(x)=(-1)^{n-1}\int_{x_0}^x\frac{(t-x)^{n-1}}{(n-1)!}f^{(n)}(t) dt=(-1)^{n-1}f^{(n)}(\xi)\int_{x_0}^x\frac{(t-x)^{n-1}}{(n-1)!} dt\\=(-1)^n\frac{(x_0-x)^n}{n!}f^{(n)}(\xi)=\frac{(x-x_0)^n}{n!}f^{(n)}(\xi),\quad\text{for some }\xi\in(x\wedge x_0,x\vee x_0)$$

where in the second step I used the mean value theorem for integrals because the polynomial $(t-x)^{n-1}$ doesnt change of sign in $[x\wedge x_0,x\vee x_0]$.

NOTE: here the symbolds "$\wedge$" and "$\vee$" means "minimum" and "maximum" respectively.