Prove $ frac{x}{2x-arctan x } $ continuous
$begingroup$
$$ f(x) = frac{x}{2x-arctan x } $$
Prove that $ f(x) $ isn't continuous at only one single point.
Obviously $ x = 0 Rightarrow 2x-arctan x = 0 $ and therefore it isn't continuous at $ x = 0 $ but how can I show that the are no other points?
I was thinking about the graph of $ g(x) = arctan x $ which gravitate between $ -pi/2 $ to $ pi/2 $ and so the only option for the $ 2x - arctan x = 0 $ only if $ x = 0 $.
but that's not enough as a proof. I guess I should assume there is another point and get contradiction?
calculus
asked Jan 6 at 19:08
bm1125bm1125
67816
$endgroup$
add a comment |
$begingroup$
$$ f(x) = frac{x}{2x-arctan x } $$
Prove that $ f(x) $ isn't continuous at only one single point.
Obviously $ x = 0 Rightarrow 2x-arctan x = 0 $ and therefore it isn't continuous at $ x = 0 $ but how can I show that the are no other points?
I was thinking about the graph of $ g(x) = arctan x $ which gravitate between $ -pi/2 $ to $ pi/2 $ and so the only option for the $ 2x - arctan x = 0 $ only if $ x = 0 $.
but that's not enough as a proof. I guess I should assume there is another point and get contradiction?
calculus
asked Jan 6 at 19:08
bm1125bm1125
67816
$endgroup$
add a comment |
$begingroup$
$$ f(x) = frac{x}{2x-arctan x } $$
Prove that $ f(x) $ isn't continuous at only one single point.
Obviously $ x = 0 Rightarrow 2x-arctan x = 0 $ and therefore it isn't continuous at $ x = 0 $ but how can I show that the are no other points?
I was thinking about the graph of $ g(x) = arctan x $ which gravitate between $ -pi/2 $ to $ pi/2 $ and so the only option for the $ 2x - arctan x = 0 $ only if $ x = 0 $.
but that's not enough as a proof. I guess I should assume there is another point and get contradiction?
calculus
asked Jan 6 at 19:08
bm1125bm1125
67816
$endgroup$
$$ f(x) = frac{x}{2x-arctan x } $$
Prove that $ f(x) $ isn't continuous at only one single point.
Obviously $ x = 0 Rightarrow 2x-arctan x = 0 $ and therefore it isn't continuous at $ x = 0 $ but how can I show that the are no other points?
I was thinking about the graph of $ g(x) = arctan x $ which gravitate between $ -pi/2 $ to $ pi/2 $ and so the only option for the $ 2x - arctan x = 0 $ only if $ x = 0 $.
but that's not enough as a proof. I guess I should assume there is another point and get contradiction?
calculus
calculus
asked Jan 6 at 19:08
bm1125bm1125
67816
asked Jan 6 at 19:08
bm1125bm1125
67816
asked Jan 6 at 19:08
bm1125bm1125
67816
asked Jan 6 at 19:08
bm1125bm1125
67816
asked Jan 6 at 19:08
bm1125bm1125
67816
67816
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We want to show that $2x-arctan x =0$ has only one root, $x=0$.
Let $g(x)=2x-arctan x$. Then $g'(x)= 2 - frac{1}{1+x^2} geq 1$.
Assume for a contradiction that $g$ has another root: $r$.
As $g$ is continuous and differentiable everywhere, and $g(0)=g(r)$, we can apply Rolle's theorem.
By Rolle's theorem, $g'$ has a root between $0$ and $r$.
But we found that $g' geq 1$. Contradiction. There are no other roots.
answered Jan 6 at 19:16
ThePortakalThePortakal
4,03011527
$endgroup$
add a comment |
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1 Answer
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$begingroup$
We want to show that $2x-arctan x =0$ has only one root, $x=0$.
Let $g(x)=2x-arctan x$. Then $g'(x)= 2 - frac{1}{1+x^2} geq 1$.
Assume for a contradiction that $g$ has another root: $r$.
As $g$ is continuous and differentiable everywhere, and $g(0)=g(r)$, we can apply Rolle's theorem.
By Rolle's theorem, $g'$ has a root between $0$ and $r$.
But we found that $g' geq 1$. Contradiction. There are no other roots.
answered Jan 6 at 19:16
ThePortakalThePortakal
4,03011527
$endgroup$
add a comment |
$begingroup$
We want to show that $2x-arctan x =0$ has only one root, $x=0$.
Let $g(x)=2x-arctan x$. Then $g'(x)= 2 - frac{1}{1+x^2} geq 1$.
Assume for a contradiction that $g$ has another root: $r$.
As $g$ is continuous and differentiable everywhere, and $g(0)=g(r)$, we can apply Rolle's theorem.
By Rolle's theorem, $g'$ has a root between $0$ and $r$.
But we found that $g' geq 1$. Contradiction. There are no other roots.
answered Jan 6 at 19:16
ThePortakalThePortakal
4,03011527
$endgroup$
add a comment |
$begingroup$
We want to show that $2x-arctan x =0$ has only one root, $x=0$.
Let $g(x)=2x-arctan x$. Then $g'(x)= 2 - frac{1}{1+x^2} geq 1$.
Assume for a contradiction that $g$ has another root: $r$.
As $g$ is continuous and differentiable everywhere, and $g(0)=g(r)$, we can apply Rolle's theorem.
By Rolle's theorem, $g'$ has a root between $0$ and $r$.
But we found that $g' geq 1$. Contradiction. There are no other roots.
answered Jan 6 at 19:16
ThePortakalThePortakal
4,03011527
$endgroup$
We want to show that $2x-arctan x =0$ has only one root, $x=0$.
Let $g(x)=2x-arctan x$. Then $g'(x)= 2 - frac{1}{1+x^2} geq 1$.
Assume for a contradiction that $g$ has another root: $r$.
As $g$ is continuous and differentiable everywhere, and $g(0)=g(r)$, we can apply Rolle's theorem.
By Rolle's theorem, $g'$ has a root between $0$ and $r$.
But we found that $g' geq 1$. Contradiction. There are no other roots.
answered Jan 6 at 19:16
ThePortakalThePortakal
4,03011527
answered Jan 6 at 19:16
ThePortakalThePortakal
4,03011527
answered Jan 6 at 19:16
ThePortakalThePortakal
4,03011527
answered Jan 6 at 19:16
ThePortakalThePortakal
4,03011527
4,03011527
add a comment |
add a comment |
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