Question

Mean value theorem. Help!

Answer 1

For the function: f(x)=^3-2x^2-4x-1 on interval [0,1]
a) verify this function satisfies the conditions of the mean value theorem
b) find the values of c that satisfy the conclusion of the mean value theorem

Answer 2

ok so this is the only one where I haven't worked it out yet. I can't quite remember how to do the mean value theorem.

Answer 3

you need to fist check to see if the conditions are met
that is check to see if f is continuous on [0,1]
and differentiable on (0,1)

Answer 4

is f(x)=x^3-2x^2-4x-1

Answer 5

"Because it's a polynomial, we know that it is continuous."
is the first thing I should say, yeah?

Answer 6

ooh oh then I put
f(0)
and f(1)
right?!

Answer 7

im just trying to boost my smart score. ignore me comment

Answer 8

yes poly are continuous and differentiable everywhere

yes and so we can apply mvt
so you will need to find the following first:
f(0)
f(1)
f'(x)

Answer 9

ok, give me just a moment :)

Answer 10

MVT says:
IF f is continuous on [a,b] and diff on (a,b) then there exist c in (a,b)
such that \[f'(c)=\frac{f(b)-f(a)}{b-a} \\ \text{ we have } \\ f'(c)=\frac{f(1)-f(0)}{1-0} \\ \]

Answer 11

this will be easier to see how to solve for c once you find f(1) and f(0) and f'(x)

Answer 12

you tell me if you are confused

Answer 13

okay :) nah i'm not confused yet! i am just writing it all out haha sorry.

Answer 14

ok take your time then

Answer 15

eeem..
f(0)=-1
f(1) = -6
...?

Answer 16

I feel like one of those should be positive o.o

Answer 17

i got that those same values

Answer 18

oh ok. Typically there is one positive one :| hm hm

Answer 19

\[f'(c)=\frac{f(1)-f(0)}{1-0} \\ f'(c)=\frac{-6-(-1)}{1-0}\]

Answer 20

f'(x) =3x^2-4x-4

Answer 21

f'(c)=3c^2-4c-4

Answer 22

\[3c^2-4c-4=\frac{-6-(-1)}{1-0}\]

Answer 23

this is quadratic equation

Answer 24

find c such that c is in the interval (0,1)

Answer 25

so that is the end of part a

Answer 26

no we did a a long time ago :p

Answer 27

The conditions that we needed to check off were:
Is f continuous on [0,1]?
Is f differentiable on (0,1)?
Those are the conditions

Answer 28

we are on part b when trying to solve the equation
\[f'(c)=\frac{f(1)-f(0)}{1-0}\]

Answer 29

okay

Answer 30

so now we've got that quadratic thing, and we just solve for c?

Answer 31

yep we want to solve that equation for c
where c is a number between 0 and 1 (exclusive)

Answer 32

aha! gime a moment

Answer 33

first couple of steps:
\[3c^2-4c-4=-5 \\ 3c^2-4c+1=0\]
hint that left hand side is factorable

Answer 34

1/3 and 1

Answer 35

right and 1/3 is the only one of those in (0,1)

Answer 36

so the end of your paper could look like this