Amistre64, still around?

Question

amistre64 · Answer

... is that a fat joke?...

anonymous · Answer

lol.  I'm so glad to have you to lighten up the fact that calculus stinks!  i have a piecewise function to type in here.  Let me try to do it without the equation thingy,ok?

amistre64 · Answer

.    .        .           .              .                 .                  *splat*

anonymous · Answer

-2x-3, x<or = to 2
-2x+5, x>2

Need to find where this is undefined.  How?

amistre64 · Answer

-2x-3,  x </ 2 

-2x+5,  x > 2

amistre64 · Answer

the y value is defined for all cases; there is a gap between the lines at x=2 but it is still defined there

watchmath · Answer

Let me write it down in more beautiful typesett :D
$\begin{cases}-2x-3&,x\leq 2\-2x+5&,x>2\end{cases}$

amistre64 · Answer

maybe its asking for the missing gap?

amistre64 · Answer

(-7,1)  perhaps; do we have options?

watchmath · Answer

Maybe asking where the function is discontinuous.

anonymous · Answer

Ok, here's what it is asking for:  To find where y' is undefined, find the value of x where the derivative on one side is different than the derivative on the other side.  Does that make sense to you?

anonymous · Answer

Amistre64, are you still here?

amistre64 · Answer

yes, english is my native tongue

watchmath · Answer

When the function is discontinuous at a point, then the derivative is undefined there/ So y' is undefined at x=2.

amistre64 · Answer

the derivative of both side = -2

amistre64 · Answer

the derivatives are never different, so I assume the answer is defined for all x

amistre64 · Answer

y ' = -2 at x=2

amistre64 · Answer

y = -2x+5 at x=2  its defined there by the domain (-inf,2]

amistre64 · Answer

it jumps yes; but it is defined

watchmath · Answer

That is not correct amistre64. Remember that diffferentiability implies continuity.

amistre64 · Answer

im reading up on that now :)  but does "undefined" pertain to jumps?  id like to verify your answer :)

amistre64 · Answer

the limit from the left and the limit from the right of piecewise function may be different; so the limit of the function does not exists at x=2

amistre64 · Answer

the limit exists from the left; and from the right; but the two are not the same...

watchmath · Answer

you can try to use the definition of derivative to check it.

anonymous · Answer

Thanks again!  I'm sure I'll be back!

amistre64 · Answer

i had to first figure out what the question was.. you did good watchmath :)