Question

Suppose f (x) = x^-1 on the interval [-1, 2]. Use the Mean Value Theorem, if it applies, to find all values c in the open interval (-1, 2) such that f'(c) = (f(2) - f(-1))/(2-(-1)).

Options:
c= -1/2
c= 1/2
c= 9/2
Mean Value Theorem does not apply

Answer 1

HI!!

Answer 2

your function is not defined on all real numbers is it?

Answer 3

what is the domain of \[f(x)=\frac{1}{x}\]?

Answer 4

Hi! :)
That was the entire question, I'm not sure what the domain of f(x) = 1/x is :o

Answer 5

i don't believe you

Answer 6

i mean i believe it was the entire question, i don't believe you do not know where \(\frac{1}{x}\) is undefined

Answer 7

Ohhh sorry I didn't understand what you were asking :)
That would be when x=0

Answer 8

right it is not defined at \(x=0\)

Answer 9

so if you read the hypothesis of the mvt you will see that the function has to be continuous on the closed interval \([a,b]\) and differentiable in the open interval \((a,b)\)

Answer 10

is your function continuous on \([-1,2]\)?

Answer 11

No it would be closed :)

Answer 12

??

Answer 13

"closed" is a property of the interval, not of the function

Answer 14

Er sorry it wouldn't be continuous

Answer 15

the question is, is \(f(x)=\frac{1}{x}\) continuous on the entire interval \([1,2]\)?

Answer 16

right, it is not

Answer 17

because it is not even defined at 0, so it certainly cannot be continuous there, let alone differentiable

Answer 18

therefore the mean value theorem says nothing in this case, because it does not fit the hypothesis

Answer 19

Ohhhhh that makes so much more sense! Thank you very much! :)

Answer 20

\[\color\magenta\heartsuit\]

Answer 21

Have a great day and a Happy New Year! :)

Answer 22

this picture is worth 1000 word :)