Question

For f(x) = x^2 – 3x + 1, find all the values of c on the interval (0, 1) that satisfy the Mean Value Theorem. I keep getting 2???

Answer 1

I do f(b) - f(a) / b - a so I get -1-1/1-0 so I get -2/1? so I get -2? I know this is wrong but its the answer I keep getting

Answer 2

?

Answer 3

can not be done because the endpoints are not defined. Unless you meant [0,1]

Answer 4

this is how it was written....

Answer 5

Assume that the interval [0,1], you have take int on that interval

Answer 6

then i would say the hypotheses does not meet, therefor, can not be done

Answer 7

int?

Answer 8

the values of "x" between 0. .... 1

to see around what point it changes sign

Answer 9

between .32 and .39

Answer 10

using that table, yes

Answer 11

okay sooo now what?

Answer 12

now .... you found them =)

Answer 13

the idea is to make a table of small values in between, to spot the sign change

Answer 14

okay so my options are
c = 1
c = - 0.5
c = 0.5
c = 0

Answer 15

hmmm are you supposed to pick one?

Answer 16

yeah... that's how I knew my -2 answer was wrong

Answer 17

then it's no... that one,.... dunno

Answer 18

$$
f(x) = x^2 – 3x + 1\\
f'(x)=2x-3\\
$$
MVT, says that there exists a c such that
$$
f'(c)=\cfrac{f(b)-f(a)}{b-a}
$$
Let's find that c.
The average slope between b and a is
$$
\cfrac{-1-1}{1-0}=-2
$$
So then we need \(f'(c)=-2\):
$$
f'(c)=2c-3=-2\\
\implies 2c=-2+3\\
\implies c=\cfrac{1}{2}
$$

That it!