anonymous
  • anonymous
use the definition of continuity to determine whether f is continuous at a.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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perl
  • perl
I can't read that because i don't have microsoft word. one moment
anonymous
  • anonymous
ok

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perl
  • perl
Dowloading openoffice
anonymous
  • anonymous
ok thank you so much for helping me
perl
  • perl
apparently the file to download is huge
anonymous
  • anonymous
f(x)= { 2-x if x<1 { 1 if x=1 { x^2 if x>1 a=1
perl
  • perl
for continuity we need to check the points where the function changes, the rest of the function is continuous in its respective parts.
anonymous
  • anonymous
ok how do we do that
perl
  • perl
Definition of continuity: Left hand limit = right hand limit = value at the point
perl
  • perl
We can write this in symbols.
anonymous
  • anonymous
you write it with a little apostrophe thing right
perl
  • perl
Definition of continuity: $$ \Large \lim_{x \to a} f(x) = f(a)$$But this implies $$ \Large \lim_{x \to a^{+}} f(x) = \Large \lim_{x \to a^{-}} f(x) = f(a) $$
perl
  • perl
So we need to check the left limit and right limits and make sure it equals to the value at that point.
perl
  • perl
$$ \rm \Large{f(x)= \begin{cases} 2-x ~~~if~ x<1\\ 1 ~~~~~~~~ ~\rm ~if~ x=1\\ x^2 \rm ~~~~~~~~if~ x>1 \end{cases} } $$
anonymous
  • anonymous
for all the x's i replace them with a 1?
perl
  • perl
\[ \rm \Large{f(x)= \begin{cases} 2-x ~~~if~ x<1\\ 1 ~~~~~~~~ ~\rm ~if~ x=1\\ x^2 \rm ~~~~~~~~if~ x>1 \end{cases} } \] \[ \Large \lim_{x \to 1^{+}} f(x) = \\~\\ \Large \lim_{x \to 1^{-}} f(x) = \\~\\ \Large f(1) = \]
perl
  • perl
yes thats correct. and we need to check that all three of those are equal . thats the continuity condition
anonymous
  • anonymous
so the answer is 1 for all of them.
perl
  • perl
\[ \rm \Large{f(x)= \begin{cases} 2-x ~~~if~ x<1\\ 1 ~~~~~~~~ ~\rm ~if~ x=1\\ x^2 \rm ~~~~~~~~if~ x>1 \end{cases} } \] \[ \Large \lim_{x \to 1^{-}} f(x) =\lim_{x \to 1^{-}} 2-x = 2-1 = 1 \\~\\ \Large \lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} x^2 = 1^2 = 1 \\~\\ \Large f(1) = 1 \]
anonymous
  • anonymous
Thank You So Much,can you help me with one more similar to this one.
perl
  • perl
ok
anonymous
  • anonymous
determine for what numbers, if any, the given function is discontinuous. f(x)={x+7 if x is less than or equal to 0 {7 if 0 is 3
anonymous
  • anonymous
ok whats the difference between the first set and the second set. (not the questions but what you did)
perl
  • perl
there are two potential points of discontinuity, where the function abruptly changes. it changes at x =0 and x = 3. so we have to test those two parts independently
anonymous
  • anonymous
ok so how do we do that
perl
  • perl
We can look at the graph that might make it more clear https://www.desmos.com/calculator/pcjrrf5xdh
perl
  • perl
edit* \[ \rm \Large{f(x)= \begin{cases} x+7 ~~~if~ x\leq0\\ 7 ~~~~~~~~ ~\rm ~if~ 0 < x\leq 3\\ x^2-1 \rm ~~if~ x>3 \end{cases} } \] \[ \Large \lim_{x \to 0^{-}} f(x) =\lim_{x \to 0^{-}} x+7 = 0+7 = 7 \\~\\ \Large \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} 7 = 7 \\~\\ \Large f(0) =0+7 = 7 \] \[ \Large \lim_{x \to 3^{-}} f(x) =\lim_{x \to 3^{-}}7 = 7 \\~\\ \Large \lim_{x \to 3^{+}} f(x) = \lim_{x \to 3^{+}} x^2-1 = 3^2-1 = 8 \\~\\ \Large f(3) = 7 \]
perl
  • perl
from the graph and from the continuity definition we see that there is a point of discontinuity at x = 3. x=0 is fine
anonymous
  • anonymous
so all the question was asking for was two points
perl
  • perl
Before we do any work the function is potentially discontinuous at the two points x=0 and x = 3 because the function changes suddenly at those points. But after analyzing the function graphically, or using the limit above, we see that it is actually only discontinuous at one point. x= 3

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