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Mathematics 15 Online
OpenStudy (anonymous):

Amistre64, still around?

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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?

OpenStudy (amistre64):

. . . . . . *splat*

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (watchmath):

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

OpenStudy (amistre64):

maybe its asking for the missing gap?

OpenStudy (amistre64):

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

OpenStudy (watchmath):

Maybe asking where the function is discontinuous.

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

Amistre64, are you still here?

OpenStudy (amistre64):

yes, english is my native tongue

OpenStudy (watchmath):

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

OpenStudy (amistre64):

the derivative of both side = -2

OpenStudy (amistre64):

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

OpenStudy (amistre64):

y ' = -2 at x=2

OpenStudy (amistre64):

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

OpenStudy (amistre64):

it jumps yes; but it is defined

OpenStudy (watchmath):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (watchmath):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

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