1. ## Mean value theorem

I'm having some trouble with this question

Use the mean value theorem to show that:

$\displaystyle 1)\ \frac{4}{27}<3-23^ \frac{1}{3}< \frac{3}{16}$

Would you start of like this:

$\displaystyle \frac{f(27)-f(23)}{27-23}= f'(c),\ c \in (23,27)$

$\displaystyle Let\ f(x)=x^ \frac{1}{3}$

$\displaystyle \frac{27^ \frac{1}{3} - 23^ \frac{1}{3}}{27-23} = \frac{1}{3c^ \frac{2}{3}}$

$\displaystyle \frac{3- 23^ \frac{1}{3}}{4}= \frac{1}{3c \frac{2}{3}}$

$\displaystyle 23<c<27$

$\displaystyle 23^ \frac{2}{3}<c^ \frac{2}{3}< 27^ \frac{2}{3}$

$\displaystyle 3* 23^ \frac{2}{3}<3c^ \frac{2}{3}< 3*9$

$\displaystyle \frac{1}{3* 23^ \frac{2}{3}}> \frac{1}{3c^ \frac{2}{3}}> \frac{1}{27}$

From here on, I don't know what to do?

2. Originally Posted by acevipa
I'm having some trouble with these two questions:

$\displaystyle 1)\ \frac{4}{27}<3-23^ \frac{1}{3}< \frac{3}{16}$

For 1) would you start off like this. I don't really know what to do after.

$\displaystyle \frac{27^ \frac{1}{3}-23^ \frac{1}{3}}{27-23}= \frac{1}{3c \frac{2}{3}}$
That is meaningless, try posting the original question as set.

CB

3. Originally Posted by CaptainBlack
That is meaningless, try posting the original question as set.

CB

Sorry, this is question I'm having trouble with:

Use the mean value theorem to show that:

$\displaystyle 1)\ \frac{4}{27}<3-23^ \frac{1}{3}< \frac{3}{16}$

Would you start of like this:

$\displaystyle \frac{f(27)-f(23)}{27-23}= f'(c),\ c \in (23,27)$

$\displaystyle Let\ f(x)=x^ \frac{1}{3}$

$\displaystyle \frac{27^ \frac{1}{3} - 23^ \frac{1}{3}}{27-23} = \frac{1}{3c^ \frac{2}{3}}$

$\displaystyle \frac{3- 23^ \frac{1}{3}}{4}= \frac{1}{3c \frac{2}{3}}$

$\displaystyle 23<c<27$

$\displaystyle 23^ \frac{2}{3}<c^ \frac{2}{3}< 27^ \frac{2}{3}$

$\displaystyle 3* 23^ \frac{2}{3}<3c^ \frac{2}{3}< 3*9$

$\displaystyle \frac{1}{3* 23^ \frac{2}{3}}> \frac{1}{3c^ \frac{2}{3}}> \frac{1}{27}$

From here on, I don't know what to do?