Continuity of a function $f(x)=begin{cases} -1, & x0 end{cases}$
$begingroup$
This is a homework problem so I would prefer hints to answers.
$b in mathbb{R} $
$f(x)=begin{cases}
-1, & x<0 \
b, & x=0 \
+1, & x>0
end{cases}$
Does a number b exist so that $f(x)$ is continous?
I believe $f(x)$to be continuous for $x>0$ and $x<0$ due to the fact that if i made the function $g(x)= 1 $ $forall x > 0$ it would be continous, same for -1. but I'm not sure how to go about it for $x=0$
continuity
asked Dec 1 '18 at 14:15
WintherWinther
247
$endgroup$
add a comment |
$begingroup$
This is a homework problem so I would prefer hints to answers.
$b in mathbb{R} $
$f(x)=begin{cases}
-1, & x<0 \
b, & x=0 \
+1, & x>0
end{cases}$
Does a number b exist so that $f(x)$ is continous?
I believe $f(x)$to be continuous for $x>0$ and $x<0$ due to the fact that if i made the function $g(x)= 1 $ $forall x > 0$ it would be continous, same for -1. but I'm not sure how to go about it for $x=0$
continuity
asked Dec 1 '18 at 14:15
WintherWinther
247
$endgroup$
add a comment |
$begingroup$
This is a homework problem so I would prefer hints to answers.
$b in mathbb{R} $
$f(x)=begin{cases}
-1, & x<0 \
b, & x=0 \
+1, & x>0
end{cases}$
Does a number b exist so that $f(x)$ is continous?
I believe $f(x)$to be continuous for $x>0$ and $x<0$ due to the fact that if i made the function $g(x)= 1 $ $forall x > 0$ it would be continous, same for -1. but I'm not sure how to go about it for $x=0$
continuity
asked Dec 1 '18 at 14:15
WintherWinther
247
$endgroup$
This is a homework problem so I would prefer hints to answers.
$b in mathbb{R} $
$f(x)=begin{cases}
-1, & x<0 \
b, & x=0 \
+1, & x>0
end{cases}$
Does a number b exist so that $f(x)$ is continous?
I believe $f(x)$to be continuous for $x>0$ and $x<0$ due to the fact that if i made the function $g(x)= 1 $ $forall x > 0$ it would be continous, same for -1. but I'm not sure how to go about it for $x=0$
continuity
continuity
asked Dec 1 '18 at 14:15
WintherWinther
247
asked Dec 1 '18 at 14:15
WintherWinther
247
asked Dec 1 '18 at 14:15
WintherWinther
247
asked Dec 1 '18 at 14:15
WintherWinther
247
asked Dec 1 '18 at 14:15
WintherWinther
247
247
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
- You should be able to conclude from observing $lim_{x to 0^+} f(x)$ and $lim_{x to 0^-} f(x)$
answered Dec 1 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
I was going to post exactly this but you're faster :(
$endgroup$
– orlp
Dec 1 '18 at 14:20
$begingroup$
So this means that it is discontinous in $x=0$? if I understand correctly
$endgroup$
– Winther
Dec 1 '18 at 14:30
$begingroup$
that's correct.
$endgroup$
– Siong Thye Goh
Dec 1 '18 at 14:31
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
- You should be able to conclude from observing $lim_{x to 0^+} f(x)$ and $lim_{x to 0^-} f(x)$
answered Dec 1 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
I was going to post exactly this but you're faster :(
$endgroup$
– orlp
Dec 1 '18 at 14:20
$begingroup$
So this means that it is discontinous in $x=0$? if I understand correctly
$endgroup$
– Winther
Dec 1 '18 at 14:30
$begingroup$
that's correct.
$endgroup$
– Siong Thye Goh
Dec 1 '18 at 14:31
add a comment |
$begingroup$
Hint:
- You should be able to conclude from observing $lim_{x to 0^+} f(x)$ and $lim_{x to 0^-} f(x)$
answered Dec 1 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
I was going to post exactly this but you're faster :(
$endgroup$
– orlp
Dec 1 '18 at 14:20
$begingroup$
So this means that it is discontinous in $x=0$? if I understand correctly
$endgroup$
– Winther
Dec 1 '18 at 14:30
$begingroup$
that's correct.
$endgroup$
– Siong Thye Goh
Dec 1 '18 at 14:31
add a comment |
$begingroup$
Hint:
- You should be able to conclude from observing $lim_{x to 0^+} f(x)$ and $lim_{x to 0^-} f(x)$
answered Dec 1 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
Hint:
- You should be able to conclude from observing $lim_{x to 0^+} f(x)$ and $lim_{x to 0^-} f(x)$
answered Dec 1 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 1 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 1 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 1 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
$begingroup$
I was going to post exactly this but you're faster :(
$endgroup$
– orlp
Dec 1 '18 at 14:20
$begingroup$
So this means that it is discontinous in $x=0$? if I understand correctly
$endgroup$
– Winther
Dec 1 '18 at 14:30
$begingroup$
that's correct.
$endgroup$
– Siong Thye Goh
Dec 1 '18 at 14:31
add a comment |
$begingroup$
I was going to post exactly this but you're faster :(
$endgroup$
– orlp
Dec 1 '18 at 14:20
$begingroup$
So this means that it is discontinous in $x=0$? if I understand correctly
$endgroup$
– Winther
Dec 1 '18 at 14:30
$begingroup$
that's correct.
$endgroup$
– Siong Thye Goh
Dec 1 '18 at 14:31
$begingroup$
I was going to post exactly this but you're faster :(
$endgroup$
– orlp
Dec 1 '18 at 14:20
$begingroup$
I was going to post exactly this but you're faster :(
$endgroup$
– orlp
Dec 1 '18 at 14:20
$begingroup$
So this means that it is discontinous in $x=0$? if I understand correctly
$endgroup$
– Winther
Dec 1 '18 at 14:30
$begingroup$
So this means that it is discontinous in $x=0$? if I understand correctly
$endgroup$
– Winther
Dec 1 '18 at 14:30
$begingroup$
that's correct.
$endgroup$
– Siong Thye Goh
Dec 1 '18 at 14:31
$begingroup$
that's correct.
$endgroup$
– Siong Thye Goh
Dec 1 '18 at 14:31
add a comment |
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