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

Question

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

love_to_love_you · Answer

@sunnnystrong

sunnnystrong · Answer

So... Recall that the derivative does not exist at Sharp corners. @love_to_you... What do you think?

love_to_love_you · Answer

Gah, I'm not sure.. I'm really terrible at this all

mathmale · Answer

Sorry you feel bad about your math skills, but saying you're "really terrible" is not going to help.  Why not approach this with the attitude that you definitely can learn the material?

Can you first graph y=|x|?  Where is its "sharp corner" located?
Next, can you graph y=|x+3|?  To do this, you translate your graph of y=|x| to the left.
Same question:  Where is the "sharp corner" in this new case?

We need to focus on the given interval, (-6,3).  From x=-6 to x=location of "sharp corner," the graph is a straight line with negative slope.  From x=location of "sharp corner," graph is a straight line with positive slope.

Please consider graphing this function on the interval (-6,3).  Then try to answer the original question.