Question

f(x)=x^2+2x-1
How to prove f(x) is continuous on [0x1]?

Answer 1

f(x)=x^2+2x-1
How to prove f(x) is continuous on [0,1]?

Answer 2

Differentiate, show it's differentiable for [0,1]

Answer 3

But isn't it true that something has to be continuous in order for it to be differentiable?

You can't differentiate something with a jump discontinuity or a removable?

Answer 4

IF it's differentiable everywhere, THEN it's always continuous everywhere

The other way around is not true. |x| is continuous everywhere, not differentiable at x=0.

Answer 5

Yes, something has to be continuous in order to be differentiable.

But, that's not the same thing as: if continuous everywhere, then differentiable everywhere.

Answer 6

ie If you know a function is differentiable everywhere, you KNOW it is continuous everywhere.

If a function is continuous everywhere, you do NOT know if it's differentiable everywhere.

Answer 7

I'm getting confused by the difference between proving continuity and differentiability:

1. Proving continuity:

Lim h->0 f(c) has to exist and be finite
f(c) has to exist and be finite
Lim h-->0 f(c) = f(c)

2. How do you prove differentiability again?

Answer 8

many links to show why differentiability implies continuity:
https://www.google.com/search?q=differentiability+implies+continuity&oq=differentiability+implies+continuity&aqs=chrome..69i57j0l5.1443j0j4&sourceid=chrome&espv=210&es_sm=93&ie=UTF-8

Answer 9

2. How do you prove differentiability again?

by showing that the derivative is defined for all values in the interval

Answer 10

How do you show that specifically?

Answer 11

I know something is differentiable if Lim h->0 (f(a+h)-f(a))/h exists and is finite by Newton's, but how do you prove it will be differentiable everywhere?

Answer 12

Continuous means that the function value at every point equals the limit at that point.
Differentiable means that the limit from both sides exist and are equal to each other.

Answer 13

Well i guess one way to show it, is just to find the domain of the derivative - if it's domain is all real numbers, then it's differentiable for all reals.

@ranga shouldn't this be:
Differentiable means that the limit *of the derivative* from both sides exist and are equal to each other.

Answer 14

Since the derivative needs to have the same value on both sides of the point, and also the same value at the point

Answer 15

Yes, you are right agent0smith.

Answer 16

Thanks! I get it!

Would this have been another way?

A polynomial indicates continuity everywhere and since f(x) was a polynomial, it is continuous in [0,1]?

Answer 17

Yep, that's fine too.

Answer 18

To prove continuity, you have to show that there are no discontinuities in that interval, such as: jump (usually in a piecewise function, in which case you just check that the endpoint values agree), removable (looks like a hole in the graph), and asymptote (usually in a rational function, in which case you check the denominator for zero).

To prove differentiability, you have to show that the left-hand and right-hand limits of the derivative expression both exist and are equal.

Answer 19

(if your teacher decides it's acceptable)

Answer 20

To clarify, f is continuous at every point c in domain of f, then f is continuous. If f is continuous at every real number c, then f is continuous everywhere.

ie: 1/x is x^(-1) which is a polynomial then it is continuous but not continuous everywhere?

Answer 21

Since 0 is not part of its domain, so it is discontinuous at x=0, but f is continuous, just not continuous everywhere?

Answer 22

1/x is not a polynomial, they need to have positive integer exponents

Answer 23

Ok, so it doesn't apply!

I'm trying to prove f(x) is differentiable everywhere using Newton's

How would you do it from h-->0+ and h-->0-?

Answer 24

1/x isn't differentiable everywhere.

The derivative is -1/x^2 which is not differentiable when x=0

Answer 25

Sorry! I meant f(x) as in x^2+2x-1