anonymous
  • anonymous
need help with continuity of limits graph posted
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
top one is question below one is example
1 Attachment
anonymous
  • anonymous
hello
watchmath
  • watchmath
It seems you didn't understand what I said earlier :). All the limits and f(2)=4 in this case. So the function is continuous.

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anonymous
  • anonymous
well here are the tougher q's i dont know if she wants us to graph or not
anonymous
  • anonymous
4. Find where the Functions are discontinuous. a. F(x) = x + 3/ x2 – 9 b. F(x) = x + 6/ x2 + 36 c. F(x) = 3x - 6
watchmath
  • watchmath
Well you can try by sketching the graphs and see where the function has a hole, jump or asymptote.
anonymous
  • anonymous
i was thinking factoring the denominator
anonymous
  • anonymous
for a)
watchmath
  • watchmath
yes that's good
anonymous
  • anonymous
then it comes out to 0/x-3
anonymous
  • anonymous
i am horrible at math i will admit i dont know what to do next
watchmath
  • watchmath
when you factor you have \(\frac{x+3}{(x+3)(x-3)}\). Remember we can't divide by zero. So the function is undefined at x=-3 and x=3. Then at those points the function is discontinuous.
anonymous
  • anonymous
F(x) = x + 6/ x2 + 36 so for this its at x+6 and x-6
anonymous
  • anonymous
so at 6 and -6
watchmath
  • watchmath
But if you plug in x=-6 or 6 to x^2+6 you won't get zero.
anonymous
  • anonymous
wait its x^2 + 6
anonymous
  • anonymous
so graph is continous eveywhere
watchmath
  • watchmath
yes
anonymous
  • anonymous
its a straight line
watchmath
  • watchmath
good
anonymous
  • anonymous
F(x) = 3x - 6
anonymous
  • anonymous
same with this one?
watchmath
  • watchmath
yup
anonymous
  • anonymous
have a good one bro

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