ksaimouli
  • ksaimouli
find the domain of the function f.use limits to describe the behaviour of f at valus of x not in its domain 1/x+3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ksaimouli
  • ksaimouli
plz explain me
slaaibak
  • slaaibak
For what value of x will you divide by 0?
ksaimouli
  • ksaimouli
?

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ksaimouli
  • ksaimouli
\[1\div x+1\]
ksaimouli
  • ksaimouli
|dw:1327444420449:dw|
slaaibak
  • slaaibak
Cool. \[f(x) = {1 \over x+3}\] Where do you think will this function have a discontinuity? Aka, for what values of x is this function undefined?
ksaimouli
  • ksaimouli
-3
slaaibak
  • slaaibak
Yes. So, do you agree this function is defined for every value of x, except for -3?
ksaimouli
  • ksaimouli
yup
ksaimouli
  • ksaimouli
so ow to write end bhaviour
ksaimouli
  • ksaimouli
i am stuck with that
slaaibak
  • slaaibak
Then we can agree that the domain of the function is: \[(-\infty,-3) \cup (-3, \infty)\]
slaaibak
  • slaaibak
That means x goes from -infinity to JUST before three, meaning every value smaller than -3. and then it unites with every value larger than -3 to infinity
ksaimouli
  • ksaimouli
thx slaaibak
ksaimouli
  • ksaimouli
thank u ver much
slaaibak
  • slaaibak
For the limits: \[\lim_{x \rightarrow -3^-} { 1 \over x + 3} = -\infty\] similarly \[\lim_{x \rightarrow -3^+} { 1 \over x + 3} = \infty\] and lastly \[\lim_{x \rightarrow -3} { 1 \over x + 3} = Does NotExist\]
ksaimouli
  • ksaimouli
ok got u but what happens in this case |dw:1327445191194:dw|
ksaimouli
  • ksaimouli
i got it -2 right
slaaibak
  • slaaibak
Whichever x makes the denominator zero would make the undefined for that x. \[(-\infty, -2) \cup (-2,2) \cup (2, \infty)\] That's the domain
slaaibak
  • slaaibak
if x^2 - 4 = 0 x = -2 or x = 2

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