Question

The Mean Value Theorem can be applied to f(x)=x^2 on [-2,1]. Find all values of c in (-2,1) satisfying the conclusion of the Mean Value Theorem

Answer 1

whats the slope between the endpoints?

Answer 2

I'm not sure, I know have to derive it to f'(x)=2x and then have it f'(c)=2(c) but I don't know what to do from there

Answer 3

youve got the hardest part figured out, we simply need to compare it to the slope between the endpoints of your interval

Answer 4

how do we find the slope between 2 points?

Answer 5

f(b)-f(a)/b-a?

Answer 6

exactly

Answer 7

what are the steps though? She didn't explain it well on the notes

Answer 8

an interval is from a to b: (a,b) so we know what a and b are

the function f is stated as x^2

so, what is f(b)-f(a) and what is b-a?

Answer 9

f(-2)-f(1)/-2-1?.. and I got -3/-3

Answer 10

not sure if there is a stepping process, its a matter if finding out certain things and putting it all together

determine the slope between the endpoints.

determine the derivative of f (find f')

equate them to determine a value for c such that:

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

Answer 11

f=x^2
f' = 2x

[-2,1]
-2^2 = 4
1^2=1

4-1 = 3
1--2 = 3

3/3 = 1 is the slope between endpoints

when does 2c = 1?

Answer 12

might have thought to quickly on slope ...

b=1, a=-2
1 - 4 = -3

slope is -1

Answer 13

Ooh. So then the slope would be equal to the f'(c)?

Answer 14

And that's how we get -1/2?

Answer 15

correct

Answer 16

2c = -1, c = -1/2

Answer 17

Thank you! :)

Answer 18

youre welcome :)