Question

Find the value of c that is guaranteed by the Mean Value Theorem for the function f(x)= x^3-12x on the interval [-3,0].

Answer 1

So the `Mean Value Theorem` tells us:
Given a `secant line`, (we're using the end points of our interval for our secant line),
If the function is continuous, there MUST exist a `tangent line` somewhere in the interval that has the same slope as that secant line.

So let's first find the value of the secant line.
\[\large \frac{f(0)-f(-3)}{0-(-3)}\]
So here is the slope of our secant line.
Let's plug in some values and figure out what that slope is.

Answer 2

f(0)=0. Plugging in the -3's for f(-3) gives us something like this,
\[\large \frac{0-\left[(-3)^3-12(-3)\right]}{3}\]

Answer 3

Understand that part? :O

Answer 4

yup!

Answer 5

So what do you get for the slope of the secant line?

Answer 6

-3 I think? When you simplify it all down.
I dunno, I was trying to do the math in my head lol.

Answer 7

i got the same answer as yours, but the answers that the book is giving is different. a) -sqrt(3) b) sqrt(3) c) -2sqrt(3) d) -sqrt(5) e)sqrt(5)

Answer 8

Yes we still have a bit of work to do :)

Answer 9

So we've found the slope of the secant line.
The Mean Value Theorem is telling us that there exists a tangent line with the same slope value. In other words,\[\large f'(c)=-3\]
Where \(\large c\) is somewhere inside of our interval. That's the value we want to solve for. The derivative at a particular point represents the slope of a `tangent line`.
So this will give us what we want.

We'll first want to take the derivative of our function f.\[\large f(x)=x^3-12x\]\[\large f'(x)=?\]

Answer 10

3x^2-12

Answer 11

Ok good. From here, we'll plug in a c for our x. (This is necessary, it just follows the notation of the Theorem).\[\large f'(c)=3c^2-12\]

Since we already determined that \(\large f'(c)=-3\)

We'll set these equal to one another,\[\large 3c^2-12=-3\]
And solve for c!
If you get 2 values for c, think about which one falls inside of our interval.

Answer 12

This is not necessary* Blah typo

Answer 13

so the correct answer will be sqrt(3) :)

Answer 14

Ummm let's see. Dividing everything by 3,\[\large c^2-4=-1\]
Doing some math...\[\large c=\pm \sqrt3\]

We have to remember a positive and negative solution come out taking the root of a square.

Which one is the c value that we want?
Which one falls in our interval, [-3,0]?

Answer 15

sqrt(3)

Answer 16

no, -sqrt(3)

Answer 17

Yah the negative one falls in our interval right? Good job.