Question

PLEASE HELP! MEDAL!

Answer 1

@amistre64 @ganeshie8 @mathmath333 @pooja195

Answer 2

lets do one at a time.. ok?

Answer 3

yep, fine

Answer 4

I don't know how to find f(x)

Answer 5

a) see we need to find f(4) in order to find f'(4)
but we know the value of g(4) and g(x) can be expressed in terms of f(x)
see g(4)=f(4)/{f(4)-3}
or 3=f(4)/{f(4)-3} since g(4)=3
implies 3f(4)-9=f(4) implies 2f(4)=9 implies f(4)=9/2
and hence equation of tangent to f(x) at x=4 is
y-9/2=f'(4) { x-4}
implies y-9/2=5{x-4)

Answer 6

Oh my gosh, you're a genious

Answer 7

genius*

Answer 8

next part
b) f is differentiable for all real values and hence must be continuous too(theorem) .
So yes. f(x) is continuous at x=3

Answer 9

Ok

Answer 10

now see the 3rd one is a bit tricky..
first evaluate f(x) in terms of g(x) from the reltaion g(x)=f(x)/{f(x)-3}
You will get f(x)=3g(x)/{3g(x)-1}

Answer 11

you are given the values of g(2) and g(4) in the table.
Put these values and evaluate f(2) and f(4)

Answer 12

you will get f(2)=3/4 and f(4)=9/8
and f is continuous so by the neighbourhood property there will not exist any root of f in [2,4] since the functional values are always positive, and for root they must cut the axis of x at least once, which is not the case

Answer 13

are you following? @StudyGurl14

Answer 14

I don't get the last part

Answer 15

which one??

Answer 16

" f is continuous so by the neighbourhood property there will not exist any root of f in [2,4] since the functional values are always positive, and for root they must cut the axis of x at least once, which is not the case"

Answer 17

oh see root means where the x values where curve cuts the x-axis i.e. f(x)=0
But here f(2)>0 and f(4)>0 and the curve is continuous, so it never had f(x)=0 in [2,4] and hence the question of zero never arise

Answer 18

Oh, I see. That makes sense

Answer 19

ok lets move to part d) then

Answer 20

k

Answer 21

\[g(x)=\frac{ f(x) }{ f(x)-3 }\]

Answer 22

so \[g'(x)=\frac{ (f(x)-3)f'(x) -f'(x)f(x)}{ [f(x)-3 ]^{2}}\]

Answer 23

\[g'(x)=\frac{ -3f'(x) }{ [f(x)-3]^{2}}\]

Answer 24

now we need to find g'(2)

Answer 25

see we need to find f'(2) and f(2) in order to find g'(2)
now f'(2) is given in the table.
So here we need to find f(2).
as calculated previously,f(2)=3/4
so plug the value of f(2)=3/4 and f'(2)=3 in the expression of g'(x) and you will get g'(2)

Answer 26

Hold on a sec. I was checking what you said and try to see if I could do it myself, and I got f(2) = 3/2

Answer 27

yeah f(2) is 3/2

Answer 28

ok, cause you put f(2) = 3/4

Answer 29

I am not doing with pen and paper.. I hope you are getting the concept clear?

Answer 30

yes.

Answer 31

ok now you check the part d) and let me know if you have any problem

Answer 32

I did it myself. It works out. Thanks. :)

Answer 33

Now e

Answer 34

Oh, I got -4 for g'(2)

Answer 35

I am not checking.. you check twice and be sure..:) the method is not wrong, any error will be in the calculation. ok lets switch to e)

Answer 36

ok

Answer 37

u there @jango_IN_DTOWN Or you just trying to figure it out like I am?

Answer 38

wait i am thinking.. do you have the values of f(4),f'(4)? let me know

Answer 39

yes. f(4) = 9/2, f'(4) = 5

Answer 40

did you figure out anything yet?

Answer 41

No, I don't see why it wouldn't be differentaible. It's continuous there

Answer 42

http://www-math.mit.edu/~djk/calculus_beginners/chapter09/section02.html

Answer 43

I am looking at this link but its not helping

Answer 44

I know. Maybe it's a trick question?

Answer 45

NO IDEA

Answer 46

See the differentiabilty of g depends on f. and we see that f is differentiable at x=4, but still it says that it is not differentiable.. Well I dont get this..

Answer 47

the only way it is not differentiable if f(4)=3. but it is not the case

Answer 48

I'm just going to say that is is differentable. I'll see what my teacher says

Answer 49

wait try checking the differentiability by definition

Answer 50

??

Answer 51

\[g'(4)=\lim h->0 \frac{ g(4+h)-g(4) }{ h }\]

Answer 52

are you calculating? what did you get?

Answer 53

well, if that's a zero in front, then it's 0, isn't it?

Answer 54

no its h tends to 0

Answer 55

i got to go. Thanks for all your help.

Answer 56

oh, ok. I'll do this later.

Answer 57

if your teacher gives the solution, do let me know...

Answer 58

pretty sure it will end up being undefined though, cause h/h = 0/0 = undefined

Answer 59

@freckles will you kindly check the last part/ why not differentiable? I dont have any idea

Answer 60

hey I don't see how c can be determined
yes f(2) and f(4) are positive
but you could have...