# 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

• Please share with us what you have thought regarding this problem. – user 170039 Nov 8 '14 at 5:36
• Suggestion: $f(x) = x^{1/7}$. Observe that $f(128)=2$. – user147263 Nov 8 '14 at 6:08
• may i ask how to choose an interval for mvt? – UnusualSkill Nov 8 '14 at 7:40
• based on @Rafflesiaarnoldii hint you can choose the interval as $[128, 129]$ and apply MVT – Paramanand Singh Nov 8 '14 at 9:26
• One has $129^{1/7}>128^{1/7}=2$ without any theorem. – Christian Blatter Nov 8 '14 at 14:43

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 $2.002224705\ldots\ .$
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 $$0.0011\lt{1\over7\cdot129}\lt{1\over7\cdot129^{6/7}}\lt f'(c)\lt{1\over448}\lt0.0023$$