Question

tell whether the Mean Value theorem can be applied to f on the closed interval [a,b]. find all values of c in the open interval (a,b) such that f'(c)=f(b)-f(a)/b-a

f(x)= (2-x)^(1/2), [-7,2]

Answer 1

The question to ask here is whether or not \(f(x)\) is continuous and differentiable on the interval.

Answer 2

It has to be continuous on the closed interval [a,b]
It has to be differentiable on the open interval (a,b)
just tagging along with what wio stated

Answer 3

so then would i do the the derivative and then set it to zero, because i am lost

Answer 4

no it is best to look at the graph and determine if it is continuous and differentiable on the interval stated

Answer 5

even finding all the values in the interval too

Answer 6

that is the last part because if it is not continuous then the theorem does not apply
or if it is not differentiable then the theorem does not apply

Answer 7

so i can make a table of the values and plug the intervals into the main equation

Answer 8

no start be graphing it and then look at the interval

Answer 9

i just did and found that it is betweem the intervals

Answer 10

ok it is continuous between [-7,2]
you can take the derivative between (-7,2)

Answer 11

so MVT applies

Answer 12

you must check these two conditions before going to the last part

Answer 13

now we need to find f'(c)
and then (f(b)-f(a)) over (b-a)
set it equal to each other and solve it

Answer 14

\[\frac{ f(2)-f(-7) }{ 2-7}\]

Answer 15

find that value and set it equal to f'(c)
take the derivative of your function sub x=c
set those two equal to each other and solve for c

Answer 16

think MrGrimm left so I thought I would leave a set of instructions for him to follow

Answer 17

Are you sure it is continuous at \(2\)?

Answer 18

isn't that the endpoint and included

Answer 19

If x=2 y=0, correct?

Answer 20

But the limit

Answer 21

But does the limit at 2 also go to 0?