Continuity of a function $f(x)=begin{cases} -1, & x0 end{cases}$
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
Winther
227
add a comment |
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
Winther
227
add a comment |
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
Winther
227
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
Winther
227
asked Dec 1 '18 at 14:15
Winther
227
asked Dec 1 '18 at 14:15
Winther
227
asked Dec 1 '18 at 14:15
Winther
227
asked Dec 1 '18 at 14:15
Winther
227
227
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 Goh
99.3k1464117
I was going to post exactly this but you're faster :(
– orlp
Dec 1 '18 at 14:20
So this means that it is discontinous in $x=0$? if I understand correctly
– Winther
Dec 1 '18 at 14:30
that's correct.
– Siong Thye Goh
Dec 1 '18 at 14:31
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Goh
99.3k1464117
I was going to post exactly this but you're faster :(
– orlp
Dec 1 '18 at 14:20
So this means that it is discontinous in $x=0$? if I understand correctly
– Winther
Dec 1 '18 at 14:30
that's correct.
– Siong Thye Goh
Dec 1 '18 at 14:31
add a comment |
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 Goh
99.3k1464117
I was going to post exactly this but you're faster :(
– orlp
Dec 1 '18 at 14:20
So this means that it is discontinous in $x=0$? if I understand correctly
– Winther
Dec 1 '18 at 14:30
that's correct.
– Siong Thye Goh
Dec 1 '18 at 14:31
add a comment |
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 Goh
99.3k1464117
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 Goh
99.3k1464117
answered Dec 1 '18 at 14:19
Siong Thye Goh
99.3k1464117
answered Dec 1 '18 at 14:19
Siong Thye Goh
99.3k1464117
answered Dec 1 '18 at 14:19
Siong Thye Goh
99.3k1464117
99.3k1464117
I was going to post exactly this but you're faster :(
– orlp
Dec 1 '18 at 14:20
So this means that it is discontinous in $x=0$? if I understand correctly
– Winther
Dec 1 '18 at 14:30
that's correct.
– Siong Thye Goh
Dec 1 '18 at 14:31
add a comment |
I was going to post exactly this but you're faster :(
– orlp
Dec 1 '18 at 14:20
So this means that it is discontinous in $x=0$? if I understand correctly
– Winther
Dec 1 '18 at 14:30
that's correct.
– Siong Thye Goh
Dec 1 '18 at 14:31
I was going to post exactly this but you're faster :(
– orlp
Dec 1 '18 at 14:20
I was going to post exactly this but you're faster :(
– orlp
Dec 1 '18 at 14:20
So this means that it is discontinous in $x=0$? if I understand correctly
– Winther
Dec 1 '18 at 14:30
So this means that it is discontinous in $x=0$? if I understand correctly
– Winther
Dec 1 '18 at 14:30
that's correct.
– Siong Thye Goh
Dec 1 '18 at 14:31
that's correct.
– Siong Thye Goh
Dec 1 '18 at 14:31
add a comment |
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