anonymous
  • anonymous
Amistre64, still around?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
... is that a fat joke?...
anonymous
  • 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?
amistre64
  • amistre64
. . . . . . *splat*

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More answers

anonymous
  • anonymous
-2x-3, x2 Need to find where this is undefined. How?
amistre64
  • amistre64
-2x-3, x 2
amistre64
  • 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
watchmath
  • watchmath
Let me write it down in more beautiful typesett :D \(\begin{cases}-2x-3&,x\leq 2\\-2x+5&,x>2\end{cases}\)
amistre64
  • amistre64
maybe its asking for the missing gap?
amistre64
  • amistre64
(-7,1) perhaps; do we have options?
watchmath
  • watchmath
Maybe asking where the function is discontinuous.
anonymous
  • 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?
anonymous
  • anonymous
Amistre64, are you still here?
amistre64
  • amistre64
yes, english is my native tongue
watchmath
  • watchmath
When the function is discontinuous at a point, then the derivative is undefined there/ So y' is undefined at x=2.
amistre64
  • amistre64
the derivative of both side = -2
amistre64
  • amistre64
the derivatives are never different, so I assume the answer is defined for all x
amistre64
  • amistre64
y ' = -2 at x=2
amistre64
  • amistre64
y = -2x+5 at x=2 its defined there by the domain (-inf,2]
amistre64
  • amistre64
it jumps yes; but it is defined
watchmath
  • watchmath
That is not correct amistre64. Remember that diffferentiability implies continuity.
amistre64
  • amistre64
im reading up on that now :) but does "undefined" pertain to jumps? id like to verify your answer :)
amistre64
  • 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
amistre64
  • amistre64
the limit exists from the left; and from the right; but the two are not the same...
watchmath
  • watchmath
you can try to use the definition of derivative to check it.
anonymous
  • anonymous
Thanks again! I'm sure I'll be back!
amistre64
  • amistre64
i had to first figure out what the question was.. you did good watchmath :)

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