How to show 30^(1/3)-3 < 1/9 with help of mean value theorem?

### 1 Answer | Add Yours

You need to identify the function `f(x) = root(3) x` which is continuous over the interval [27,30] and it is differentiable (27,30).

Using the mean value theorem yields that there exists `c in (27,30)` such that:

`f'(c) = (f(30) - f(27))/(30-27)`

You need to differentiate the identified function `f(x) = root(3) x` with respect to x such that:

`f'(x) = (1/3)(x^(1/3 - 1)) => f'(x) = (1/3)(x^(-2/3)`

`f'(x) = 1/(3root(3)(x^2))`

`f'(c) = 1/(3root(3)(c^2))`

You need to evaluate`f(30) - f(27)` such that:

`f(30) - f(27) = (root(3)30 - root(3)27)`

`f(30) - f(27) = (root(3)30 - 3)`

`f'(c) = (root(3)30 - 3)/3 => 3f'(c) = (root(3)30 - 3)`

Substituting `1/(3root(3)(c^2))` for f'(c) yields:

`3(1/(3root(3)(c^2))) = (root(3)30 - 3) => 1/(root(3)(c^2))=(root(3)30 - 3) `

Hence, you need to select a value of `c in (27,30)` such that:

`28 in (27,30) => 9 < 28 => 1/9 > 1/28 >1/(root(3)(c^2)) = (root(3)30 - 3) `

**Hence, using mean value theorem yields that inequality `root(3)30 - 3 < 1/9` holds.**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes