Question

Using f(x)=(x+3)/X on [1,6] I'm supposed to find the point c guaranteed by the Mean Value Theorem.

For f(6)=(6+3)/6=3/2 and f(1)=(1+3)/1=4
This gives f '(c)=(3/2-4)/(6-1) = (-5/2)/5=-1/2

What am I supposed to do now? I took the derivative of f(x) and got -3/x^2=0
Am I supposed to do something with that or what? Any hints/help is greatly appreciated!

Answer 1

so there is at least one c in [a,b] such that f'(c) is equal to the slope between f(a) and f(b)

Answer 2

yes,

Answer 3

(x+3) (x)^-1

(x)^-1 - (x+3) (x)^-2

3/x^2 = -1/2; check to make sure your slope is correct between the end points

Answer 4

3/c^2 = k

3/k = c^2

sqrt(3/k) = c , between 1 and 6

Answer 5

I got -3/x^2

Answer 6

:) fine -3/x^2 then

Answer 7

lol

Answer 8

sqrt(6) seems to be good for x tho

Answer 9

or c

Answer 10

OK, so you don't set it equal to 0 you set it equal to the f '(c) which was -1/2 and then that gives the point on the interval

Answer 11

f' defines the slope at a single point; the mean value thrm states that there is some intermediary point between the ends that equals the slope between the ends

if that slope is not 0 then I would not equate it to zero

Answer 12

Yeah, I don't know why my original thought was to set it equal to zero, it just seems to be "the thing to do" in math. :)

Answer 13

min/max is set to 0 ;) you did fine tho

Answer 14

Just a technicality question, would the answer technically be plus or minus the sqrt(6) but it can't be minus because the interval is on [1,6] ?

Answer 15

since -sqrt6 is not in the interval, you cant really say that its a solution within the stated interval

Answer 16

Right, that's what I was getting at. If the interval was say [-6,6] then you would have to include minus sqrt(6) though right?

Answer 17

not really, the interval needs to be continuous; and this is not continuous at x=0

Answer 18

granted it works out fine in this case, but in general thats a bad idea

Answer 19

the slope of the line between -6 and 6 is prolly zero

Answer 20

OK, thanks for your help!

Answer 21

its very close to zero along -6 to 6 ... and since its discontinuous along the interval you cant guarantee a solution