Question

For f(x)= ln(x), x>e
x/e, x
12 years ago

Answer 1

f(x) is continuous and differentiable everywhere

f(x) is continuous but not differentiable at x = e

f(x) is differentiable but not continuous at x = e

f(x) is neither continuous nor differentiable

f(x) is undefined at x = e

Answer 2

those are the answer choices above. It is a piecewise exponential function from what I can tell.
I'd love some help. I've been doing math all day, and my reserves are exhausted. I will try to graph this with my calculator and see what I can figure out, but any leads or guidance would be amazing, thank you in advance!

Answer 3

Hello, thank you for taking a look at this problem? Do you have any advice?

Answer 4

Ummm so for continuity, the pieces of the function much connect.
For differentiability, the connection must be smooth.

So we gotta do a couple of limit tests.

Answer 5

Ahh crap my game is starting D::

Answer 6

I can come help later maybe :C

Answer 7

ok so the function is continuous throughout, so we should test the limit to find differentiability?
ok no problem I will try to figure this out

Answer 8

We need the limit from the left side to match the limit from the right.\[\Large\bf\sf \lim_{x\to e^{-}}f(x)\quad=\quad \lim_{x\to e^{+}}f(x)\]

Answer 9

Ok... I am a bit confused when calculating with numbers like 'e'
I'll try to set this up...

Answer 10

Ok so lim x/e would need to equal lim ln(x)?

Answer 11

Mmm yes good good good,
from the left, the function is defined to be x/e,\[\Large\bf\sf \lim_{x\to e^{-}}\frac{x}{e}\quad=\quad \lim_{x\to e^{+}}f(x)\]And from the right side it's defined to be ln(x),\[\Large\bf\sf \lim_{x\to e^{-}}\frac{x}{e}\quad=\quad \lim_{x\to e^{+}}\ln(x)\]

Answer 12

So we don't have any trouble plugging e directly in for x.\[\Large\bf\sf \frac{e}{e}=\ln(e)\]Does this hold true?
This tells us something about continuity either way.

Answer 13

Which one? :)