Question

Why is it false?

The function of f(x)=2x-3/x-1 satisfies the MVT on the interval [0,2].

Answer 1

f(2)-f(0)/2-1=-2

and is it because -2 isn't in the interval of [0,2]?

Answer 2

it's (2x-3)/(x-1) right?

Answer 3

yessir.

Answer 4

ok, notice that when x = 0, f(x) is f(0) = (2*0-3)/(0-1) = (-3)/(-1) = 3

when x = 2, f(x) is f(2) = (2*2-3)/(2-1) = (1)/(1) = 1

Answer 5

so we have these two points:

(0,3) and (2,1)

Find the slope of the line that is going through those two points for me please

Answer 6

-1

Answer 7

now the MVT states that if you have an interval [a,b] on some function f(x), then there exists some number c such that

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

Note: ( f(b) - f(a) )/(b - a) is the slope of the line you just calculated

Answer 8

yea...

Answer 9

so this means that if the MVT was true then there should be a value of c such that

f ' (c) = -1

Answer 10

can you find me such a value

Answer 11

wait.. what? I'm confused now

Answer 12

first off, what is f ' (x)

Answer 13

1/(x-1)^2

Answer 14

now solve

1/(c-1)^2 = -1

for c

Answer 15

it might help to do this

1/(c-1)^2 = -1

1 = -1(c-1)^2

-1 = (c-1)^2

(c-1)^2 = -1

Answer 16

c is a complex solution..

Answer 17

so there are no real numbers c that satisfy f ' (c) = -1

Answer 18

... can you sum all that up lol? Because I'm lost.

Answer 19

So the slope is -1... But why does it not satisfy the thm?

Answer 20

because there are no tangent lines that have a slope of -1 on [0,2]

Answer 21

the secant slope is -1, so there should be a tangent line in [0,2] that also has a slope of -1 (if the MVT is correct)

Answer 22

my gosh... I'm so lost! ;/ how do you know there are no tangent lines that have a slope of -1? is it because -1 isn't included in the interval?

Answer 23

because f ' (c) = -1 has no real solutions

Answer 24

oh so because f'(c)=-1 has no solutions it doesn't satisfy the thm.. But, if there were solutions it would satisfy the thm?

Answer 25

yes, the big reason why this theorem fails is because of the jump discontinuity at x = 1

Answer 26

@satellite73 what do you mean?

Answer 27

http://www.wolframalpha.com/input/?i=%282x-3%29%2F%28x-1%29+domain+0..2

Answer 28

you can see that the function does not exist at 1, so it sure as hell is not differentiable at 1

Answer 29

hahaha^^!!