Question

Let f(X) be continuous funcion dfined for 1<=x<=3 .if f(x) takes rational values for all values of x and f(2)=10 . Determine value of f(1.5) @Physics

Answer 1

hmm

Answer 2

............

Answer 3

is this simliar to finding a constant of variation?

Answer 4

i dunno

Answer 5

it feels to be missing parts to me, but thats just a guess

Answer 6

help

Answer 7

there are in infinite number of continuous functions from 1 to 3 that could have f(2) = 10; at least in my mind

Answer 8

f(x) = 10

f(2) = 10

f(1.5) = 10

Answer 9

???

Answer 10

i know right, thats how i feel.

it almost seems like an initial value question

Answer 11

how u got hat?

Answer 12

simple, i defined a function f(x) = 10 that is continuous from 1 to 3 and found values that for x=2 and x=1.5

Answer 13

y it cant be any othr value????

Answer 14

it can be anything it wants to be; it just so happens that this is how i defined the function.... given no other parameters it fits

Answer 15

f(x) = 3x+4 is another one that will work fine

f(2) = 10

f(1.5) = 8.5

Answer 16

oh so there is no fixed answer?

Answer 17

none that i can see with the given information.

Answer 18

any none vertical line is continuous from 1 to 3; any parabola should work as long as you aint gonna tilt it on an odd axis

Answer 19

cubics seem good

Answer 20

i hav some more questions plz help

Answer 21

k?

Answer 22

i cant read your mind; you actually have to print something out :)

Answer 23

k

Answer 24

let f be a real fn satisfying f(x+y+z)=f(x)f(y)f(z) for all real x ,y ,zif f(2)=4 and f'(0)=3 find f(0) and f'(2)

Answer 25

then f(x+y+z) - f(x)f(y)f(z) = 0 might be a good start

Answer 26

so

Answer 27

f(2) = 4 would suggest to me that x+y+z=4 since the function doesnt seem to be defined to well at the moment

Answer 28

err; rather x+y+z = 2

Answer 29

.....

Answer 30

perhaps we might accomodate this:

[f(x+y+z) - f(x)f(y)f(z)]' = f'(x+y+z) (x'+y'+z')
- ( f'(x)x' f(y)f(z)+f(x)f'(y)y' f(z)+f(x)f(y)f'(z)z' )

but thats a bit confusing to me

Answer 31

youd prolly need zarkon to sort thru it

Answer 32

Just a note.... Your answer for the initial question is correct, but there aren't an infinite number of functions that work. The question specifies that f takes only rational values, and the real line is dense with irrationals, which means that between any two rational numbers there's an infinite number of irrationals. It follows that if the function is continuous, and it doesn't take on any irrational values, it must be constant.

Answer 33

yay!! my guess was correct, somehow :)