# let f(X) = 6x^5/2. find the value c guranteed by the mean value theorem on the interval [1,4]

### 1 Answer | Add Yours

You need to remember the defnition of mean value theorem for a function `f(x), ` continuous over its domain `[a,b], ` such that:

`f'(c) = (f(b) - f(a))/(b - a)`

You need to identify the interval `[a,b] = [1,4]` and to substitute 1 for a and 4 for b in equation `f'(c) = (f(b) - f(a))/(b - a), ` such that:

`f'(c) = (f(4) - f(1))/(4 - 1) => f'(c) = 6(4^5 - 1^5)/3`

`f'(c) = 2046`

You need to differentiate the function with respect to x, such that:

`f'(x) = (6x^5/2)' => f'(x) = (6*5*x^4)/2 => f'(x) = 15x^4`

You need to substitute c for x to evaluate `f'(c), ` such that:

`f'(c) = 15c^4`

`15c^4 = 2046 => c^4 = 2046/15 => c^4= 682/5 => c= root(4)(682/5) ~~ 3.41 in [1,4]`

**Hence, evaluating the value of `c in [1,4], ` using the mean value theorem yields `c = root(4)(682/5).` **