Question

decide if the functions are differentiable at x=0 try zooming in on a graphing calculator, or calculating the derivative f'(0) from the definition

f(x)= {xsin(1/x)+x for x doesnt = 0
{0 for x=0

Answer 1

In a very general sense, a function is differentiable at a given point if it has a well defined slope at that point (If it has no cusps or "holes"). So it should be "smooth". And as far as the derivative of the function goes, you need to specify what the function actually is.

Answer 2

not sure what that tells me :o
I am lost on this one

Answer 3

you could try to actually take the derivative using the definition

Answer 4

\[f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}\]
\[\lim_{h\to 0}\frac{h\sin(\frac{1}{h})+h-0}{h}\]
\[\lim_{h\to 0}\sin(\frac{1}{h})+1\] and this limit does not exist because
\[\sin(\frac{1}{h})\] ossilates wildly as h goes to zero (as 1/h goes to infinity)

Answer 5

*oscillates