Using Mean value theorem to prove the inequality $1.995<129^{1/7}<2.005$ How can I prove the following inequality using mean value theorem?
$$1.997<129^{1/7}<2.003$$
Progress
 A: We shall use the MVT in the following form: If $f$ is differentiable on the interval $[0,x]$  then there is a $\xi\in\ ]0,x[\ $ with
$$f(x)=f(0)+x\>f'(\xi)\ .\tag{1}$$
Apply this to the function
$$f(x):=(1+x)^{1/7}\qquad(x\geq0)$$
and obtain
$$f(x)=1+x\cdot {1\over 7}(1+\xi)^{-6/7}\leq1+{x\over7}\ .$$
It follows that
$$129^{1/7}=2\left(1+{1\over 128}\right)^{1/7}\leq2\left(1+{1\over 7\cdot 128}\right)<2.002233\ .$$
On the other hand, from $129\left(1-{1\over129}\right)=128$ we get
$$129^{1/7}=2\left(1-{1\over 129}\right)^{-1/7}\ .\tag{2}$$
This time we apply $(1)$ to the function
$$f(x):=(1-x)^{-1/7}\qquad(0\leq x<1)$$
and obtain
$$f(x)=1+x\cdot{1\over7}(1-\xi)^{-8/7}\geq1+{x\over7}\ .$$
From $(2)$ it then follows that
$$129^{1/7}\geq 2\left(1+{1\over 7\cdot 129}\right)>2.002214\ .$$
The true value is $2.002224705\ldots\ .$
A: Let $f(x)=x^{1/7}$, and note that $f(128)=2$.  Note also that $f'(x)={1\over7}x^{-6/7}={1\over7f(x)^6}$.  By the Mean Value Theorem,
$${f(129)-f(128)\over129-128}=f'(c)$$
for some $128\lt c\lt 129$.  Thus $129^{1/7}=2+f'(c)$ with $|f'(c)|={1\over7f(c)^6}\lt{1\over7\cdot2^6}={1\over7\cdot64}\lt{3\over1000}$, since $1000\lt21\cdot64$.
Remark:  The inequality can, of course, be strengthened since in fact 
$$0.0011\lt{1\over7\cdot129}\lt{1\over7\cdot129^{6/7}}\lt f'(c)\lt{1\over448}\lt0.0023$$
