Question

The functions fails to be differentiable at x=0. Tell whether the problem is a corner, a cusp, a vertical tangent, or a discontinuity.

y=Cuberoot(|x|)

Is this a vertical tangent or a corner ..?

Answer 1

corner since all values for x will tun into positive (ie y will always be positive values), to prove it, just take the derivative which is 3(|x|)^2, plug in any number and u find out that u get all positives

Answer 2

yes a nice corner since
\[|x|\] is a piece wise function

Answer 3

wait satellite don't we draw vertical tangents at corners?

y=|x|

there is a corner at x=0

we also say we can draw a vertical tangent line at x=0

Answer 4

?

Answer 5

infinite number of lines tangent at the corner

Answer 6

not, however, a vertical one in this case

Answer 7

you don't want to draw a horizontal tangent at 0
because that would mean f' would have value 0 there

so we draw a vertical tangent there?

Answer 8

the point is the "tangent line" is not defined in this case

Answer 9

you can draw as many as you like.

Answer 10

right since the slope of a tangent line is undefined at x=0
we can draw a vertical tangent line at x=0 to symbolize f' is undefined there

Answer 11

then what case do you say there is a vertical tangent?

Answer 12

heck no. it is not a vertical tangent.