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

is
\[f(x)=\sqrt{2-x}\] continuous on \([-7,2]\) ?

Answer 2

yes

Answer 3

hmm maybe not

Answer 4

i mean the conclusion of the mvt might be true in any case, but i don't think this function is continuous on the closed interval, as it does not even exist to the right

Answer 5

so then how can i show that on paper

Answer 6

you can just say it

Answer 7

the domain of this function is \((-\infty,2]\)
but it is continuous from the left so that might be good enough. lets see if we can find that famous \(c\)

Answer 8

we need some numbers
\[f(-7)=\sqrt{2+7}=3\]\[f(-2)=0\]
\[\frac{f(2)-f(-7)}{2-(-7)}=\frac{3}{9}=\frac{1}{3}\]

Answer 9

messed that up didn't i?

Answer 10

\[\frac{f(2)-f(-7)}{2-(-7)}=\frac{-3}{9}=-\frac{1}{3}\]

Answer 11

i think so,

Answer 12

and
\[f'(x)=-\frac{1}{2\sqrt{2-x}}\]

Answer 13

last job is to solve
\[-\frac{1}{2\sqrt{2-x}}=-\frac{1}{3}\]

Answer 14

i have a question in why would you put 3 on top, 2\[2-(-7)=9\]

Answer 15

\[f(b)=f(2)=0\]
\[f(a)=f(-7)=3\]
\[f(b)-f(a)=f(2)-f(-7)=0-3=-3\]

Answer 16

ok, so that is all you need to do to solve this

Answer 17

it is numbers don't forget

Answer 18

\[\frac{f(b)-f(a)}{b-a}\] is a number
it is the slope of the secant line
in this case that number is \(-\frac{1}{3}\)

Answer 19

so then, that is what the question id asking for

Answer 20

the second part of the question is asking
"solve for \(c\)
\[-\frac{1}{2\sqrt{2-c}}=-\frac{1}{3}\]

Answer 21

then what would you plug in for c

Answer 22

you don't plug in anything
you solve that equation

Answer 23

oh so then set it to zero to find c

Answer 24

?

Answer 25

i think you are thinking way too hard
solve for \(c\) you can probably do it in your head

Answer 26

you got this?

Answer 27

im lost, solve for c, would you do the derivative or cross multiply?

Answer 28

i would start with \[2\sqrt{2-c}=3\]then square both sides

Answer 29

get
\[2-c=\frac{9}{4}\] then \[c=-\frac{1}{4}\]

Answer 30

did you divide by two

Answer 31

divide by two then square
or square and then divide by 4
either way

Answer 32

i was just about to say, did you square the square root

Answer 33

yes square both sides

Answer 34

http://openstudy.com/study#/updates/53b35a74e4b072c759d7acb3

Answer 35

lets write it all out with all the steps
there is a \(c\in (-7,2)\) with
\[f'(c)=\frac{f(2)-f(-7)}{2-(-7)}\]
\[-\frac{1}{2\sqrt{2-c}}=-\frac{1}{3}\]
\[2\sqrt{2-c}=3\]
\[\sqrt{2-c}=\frac{3}{2}\]
\[2-c=\frac{9}{4}\]
\[c=2-\frac{9}{4}=-\frac{1}{4}\]

Answer 36

ok thank you, im sorry it just didnt see it for some reason