# Mean Value and Rolle's Theorem

1. Feb 16, 2013

### Torshi

1. The problem statement, all variables and given/known data

1.) (MVT) f(x) = 2x^3-6x^2-48x+4 on interval [4,9]
By the Mean Value Theorem, we know there exists a c in the open interval (-4,9) such that f'( c) is equal to this mean slope. For this problem, there are two values of c that work. The smaller one is __________ and larger one is __________?

2.) Rolles Theorem: Find all numbers c that satisfy the conclusion of Rolle's Theorem for the following function and interval. Enter the values in increasing order and enter N in any blanks you don't need to use. There are 4 values for 8sin((pi*x)), [-1,1]

2. Relevant equations
1.) f(b)-f(a)/b-a

2.) None needed.

3. The attempt at a solution

1.) I found the slope by plugging in the numbers from the interval and using the equation I posted getting 44 which is correct. The problem I'm having is trying to find the two "c" values. I've done other problems like this, but can't figure this one out.
I've set the f'(x) 6x^2-12x-48 = 44(slope) I've tried the quadratic and it didn't work.
6x^2-12x-92
(-b) +/- √b^2-4ac / 2a
I got x/- √2352 which is wrong.

2.) I know that in order to proceed I need the derivative to be = 0 or undefined. But can't solve this one

2. Feb 16, 2013

### Ray Vickson

No, the formula f(b)-f(a)/b-a is wrong: it means
$$f(b) - \frac{f(a)}{b} - a.$$
If you really mean
$$\frac{f(b) - f(a)}{b-a},$$ you need to use brackets, like this: (f(b)-f(a))/(b-1).

I do not intend to guess what you mean; I will assume you mean exactly what you write!

3. Feb 16, 2013

### Dick

For part 1) you were doing just fine until the quadratic formula part. "x/- √2352" can't be the whole answer. If the answer looks different from what you got there is a much simpler way to write √2352. Try and factor 2352. For the second one what did you get when you differentiated 8sin((pi*x))?

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