Question

0

Answer 1

did you say
IS continuous at x=0
not IF continuous at x=0

this is in the chapter 9 and 10 of Spivak Calculus

Answer 2

its worded correctly above as the problem

Answer 3

okay
then check the spivak calculus on chapters 9 and 10

Answer 4

i dont have that text

Answer 5

Can you think of a counterexample?

Answer 6

I can think of one

Answer 7

no
if f(x) is continuous at x=a and g(x) is continuous at x=a then the composite function
f(g(x)) is continuous at x=a

Answer 8

there is only one type of example when everywhere is continuous but not one part is differentiable

Answer 9

Let f(x)=1/x and g(x)=x^2

Answer 10

x^2/x is not continuous at zero.

Answer 11

well it becomes x

Answer 12

No it doesn't. you may only cancel if x is not equal to zero. The function x^2/x is not defined at zero.

Answer 13

@nincompoop that has nothing to do with this question.

Answer 14

so because the product of f(x)*g(x) is continuous at 0 then both must be continuous on 0, is this simple multiplication theory or what?

Answer 15

That is not true. Let f(x) be a piecewisely defined function: f(x) = 0 for x<=0 and f(x) = 1 for x > 0.

Let g(x) = 0.

f(x)*g(x) = 0, which is continuous at zero, but clearly f(x) is not continuous at zero. Therefore the statement is false.

Answer 16

Ohh i forgot abt piecewise functions -_-

Answer 17

roger, that makes sense, ty

Answer 18

is it continuous at x=0 with f(x) = 1/x ?

Answer 19

as long as g(x) = 0, yes.

Answer 20

Is there any other example besides piecewise examples?

Answer 21

when they have different domains LOL

Answer 22

when they have different domains LOL

Answer 23

the last question was pointed at the
x^2/x

Answer 24

x^2/x is not continuous at x = 0 because it is not defined at x = 0.

Answer 25

1/x where x = 0 is not defined so I don't know how that is

Answer 26

EDIT: You're right, it's not defined at zero.

Answer 27

Ok coool thanks :D

Answer 28

and when one says < = it is really <

Answer 29

What do you mean?

Answer 30

don't ask me.... this is something I came across with and I am still trying to absorb it

Answer 31

why did you say it if you don't know what it means?

Answer 32

Nin please delete the statement :P

Answer 33

because how can it be an inequality if there's a point where it is equal

Answer 34

yeah... absorb it too

Answer 35

it is the same concept as increasing and strictly increasing

Answer 36

What? That has nothing to do with it. The function f(x) = 0 if x <= 0 and is equal to 1 if x > 0 is just the way I defined the function.

Answer 37

http://isites.harvard.edu/fs/docs/icb.topic484367.files/MTI_PracticeConceptualQs_Math1aF08.pdf
have fun learning @johnny101