Question

Identify the points at which f(x) is differentiable.
f(x) = |x + 3|; (-6, 3)

Answer 1

what grade are u in.

Answer 2

10th

Answer 3

same what math are u taking

Answer 4

Pre-cal

Answer 5

oh wow u are really smart.

Answer 6

lol not really

Answer 7

yeah im in geometry.

Answer 8

ah cool

Answer 9

@confluxepic can you help?

Answer 10

hmm what's the (-6,3? anyway?

Answer 11

see if this will help u https://www.sagemath.org/calctut/differentiability.html

Answer 12

an absolute value graph, will have an "abrupt shfit or turn", namely a "cusp" at the vertex
so. is the only place it's not differentiable at

Answer 13

. ?

Answer 14

\(\huge ?\)

Answer 15

so if I graph it the shift is the answer?

Answer 16

ok if that did not help u ignore me

Answer 17

ahemm the "cusps" or the vertex in this case is the only place is NOT differentiable
is differentiable everywhere else

Answer 18

It's loading @gluttrell

Answer 19

ohh, sorry I read that wrong

Answer 20

So it's differentiable everywhere except at x = -3?

Answer 21

bear in mind
differentiable -> has a smooth line over which a tangent line can easily sit
a cusp or an abrupt edge or break, is not smooth, and thus you can get a tangent line there, and thus is not differentiable

you can pick those up by simply setting the denominator of the derivative to 0 and solve for "x"

Answer 22

sort of ur on the right track.

Answer 23

yeap, -3, 0 is the vertex, so, is not differentiable at x = -3
everywhere else, is

Answer 24

Thank you, the both of you

Answer 25

and thus you can't get a tangent line there, and thus is not differentiable

correcting typo, "can't" I meant :)

Answer 26

yw

Answer 27

\[\frac{ d }{ dx }\left| f(x) \right|=\frac{ f \left( x \right) }{ \left| f \left( x \right) \right|}f \prime \left( x \right)\]
\[\frac{ d }{ dx }\left| x+3 \right|=\frac{ x+3 }{ \left| x+3 \right| }*1=\frac{ x+3 }{ \left| x+3 \right| }\]
it is not differential where denominator |x+3|=0
or x=-3
derivative does not exist at x=-3

Answer 28

did you just have a conversation with youself?

Answer 29

The better way to answer this question is to convert the abs. val. function into a piece wise and evaluate continuity and differentiability using the definition of continuity for the function f(x) and function f ' (x).

Also abs. value functions are never differentiable at the vertex because although it is continuous, the slope function has different right and left hand limits.