Question

Pretend we have f(x)=a|x| for some value a.
Then f'(x)=a|x|/x
what is f'(0) if a=0?
Multiple Choice.
A) f'(0) doesn't exist because 0*|0|/0 is undefined.
B) f'(0)=0 because f(x)=0 when a=0

Answer 1

B. It would only be undefined if x=0. Because 0 would be in the denominator. But since a=0, we know that anything multiplied by 0 is also 0, and anything divided by 0 is 0. So yes, B is the answer.

Answer 2

You have to piecewise define the derivative, see if left and right limits are equal

Answer 3

Sorry, I read it wrong, it is undefined

Answer 4

Not really since f'(x)=a|x|/x gives both right and left derivative

Answer 5

I'm more inclined to f'(0)=0 since a=0 implies that f(x)=0 for any x

Answer 6

you have to define it for x>0 and x<0, the function doesn't exist at zero (0/0) so you need to see if the limit exists

Answer 7

for any a, then pick the subcase

Answer 8

What I think maybe is if f(x)=ag(x)
then f'(x)=ag'(x) doesn't matter if a=0
if a=0 then f(x)=0*g(x)=0 <--but I'm kinda worried about saying f(x)=0 for all x if a=0 since g'(x) might have some restrictions due to the curve of y=g(x)

Answer 9

the function exists for zero, but the limit doesn't exist at zero (correction to your statement)

Answer 10

if a =0 then you do have a trivial solution

Answer 11

well if a=0 doesn't this allow us to say f(x)=0 for any x?

Answer 12

the derivaative doesn't here, the initial fn yes, oops

Answer 13

you know what I'm pretty sure since f(x) does exist for all x
then f(x)=a|x|=0|x|=0 for all x
therefore f'(x)=0 for all x

Answer 14

that is the trivial solution, yes. But don't you have to show work?

Answer 15

that is the work

Answer 16

hmm good argument

Answer 17

also i made up this question

Answer 18

meh, ok

Answer 19

hehe! you made it up. interesting!

Answer 20

now if a was some other value then we would have f(x)=a|x|
then f'(x)=a|x|/x
and if a doesn't equal 0 then f'(0) doesn't exist

Answer 21

the question is, does it exist for any other a though?

Answer 22

how are they getting that derivative value?

Answer 23

The derivative of |x| is |x|/x

Answer 24

I would have just piecewise defined it... it's easier to see the reason the deriv doesn't exist

Answer 25

|x|/x doesn't exist when x=0

Answer 26

True, for any a f'(0) doesn't exist

Answer 27

Oh, I don't consider that the derivative, I guess that's why.

Answer 28

@xapproachesinfinity I did ask the question because I have seen similar questions asked on OS.

Answer 29

@FibonacciChick666 there is nothing wrong with writing it as a piecewise function

Answer 30

both ways look pretty

Answer 31

they are the same in the end, but then it is easier to confuse the jump singularity at 0.

Answer 32

I see!

Answer 33

also, the second derivative, does exist and I'm not sure how that fn would do with the quotient rule

Answer 34

the second derivative will still have problem in 0?

Answer 35

nope, because the limit exists, it will equal zero

Answer 36

eh right! my attention went to something else

Answer 37

f(x)=a|x|=0 if a=0 => f'(x)=0 =>f'(0)=0
f(x)=a|x| for a not 0=>f'(x)=a|x|/x =>f'(0) does not exist
--
this is my summary for this problem...
I'm cool with this. Is this what everyone got back from this discussion? My choice B then would be right.

Answer 38

if f(x)=|x|
f''(x)=0 if x does not equal 0

Answer 39

f''(0) does not exist if f(x)=|x|

Answer 40

yes, B would be correct, and hm are you sure, becasue we are working with limits, I think it does