Differentiability and Continuity on an open interval
Let $f:[0,infty)tomathbb{R}$ be defined by: $$begin{cases}&xsin(frac{1}{x}) , , &text{if}, , x > 0\
& 0, &text{if} , , x = 0end{cases}$$
Show that $f$ is continuous on $[0,infty)$ and differentiable on $(0,infty)$. Also, show that $f$ has no local maximum or minimum in the endpoint $x = 0$ of the domain of $f$.
I can manage to prove continuity on a single point using the epsilon-delta technique, although the intervals here were a surprise, how do you go about proving such a thing? And any hints about the second part of the problem would also be appreciated.
real-analysis calculus derivatives proof-writing
asked Dec 8 at 9:12
kareem bokai
344
add a comment |
Let $f:[0,infty)tomathbb{R}$ be defined by: $$begin{cases}&xsin(frac{1}{x}) , , &text{if}, , x > 0\
& 0, &text{if} , , x = 0end{cases}$$
Show that $f$ is continuous on $[0,infty)$ and differentiable on $(0,infty)$. Also, show that $f$ has no local maximum or minimum in the endpoint $x = 0$ of the domain of $f$.
I can manage to prove continuity on a single point using the epsilon-delta technique, although the intervals here were a surprise, how do you go about proving such a thing? And any hints about the second part of the problem would also be appreciated.
real-analysis calculus derivatives proof-writing
asked Dec 8 at 9:12
kareem bokai
344
What does the phrase "local maximum or minimum in the endpoint" mean?
– BigbearZzz
Dec 8 at 10:10
According to the definition in the book am using, "A local maximum (minimum) can occur as an end point of the domain of $f$: a point $x$ in the domain of $f$ that does not belong to an open interval lying completely inside the domain". Yeah am equally as confused too :)
– kareem bokai
Dec 8 at 10:19
I don't quite understand what you want to prove here. Perhaps you want to show that $x=0$ is neither a local maximum nor a local minimum?
– BigbearZzz
Dec 8 at 10:21
Yeah, I think the way they formulated the question is horrible, we just have to show that $x = 0$ is neither a local max or min
– kareem bokai
Dec 8 at 10:25
add a comment |
Let $f:[0,infty)tomathbb{R}$ be defined by: $$begin{cases}&xsin(frac{1}{x}) , , &text{if}, , x > 0\
& 0, &text{if} , , x = 0end{cases}$$
Show that $f$ is continuous on $[0,infty)$ and differentiable on $(0,infty)$. Also, show that $f$ has no local maximum or minimum in the endpoint $x = 0$ of the domain of $f$.
I can manage to prove continuity on a single point using the epsilon-delta technique, although the intervals here were a surprise, how do you go about proving such a thing? And any hints about the second part of the problem would also be appreciated.
real-analysis calculus derivatives proof-writing
asked Dec 8 at 9:12
kareem bokai
344
Let $f:[0,infty)tomathbb{R}$ be defined by: $$begin{cases}&xsin(frac{1}{x}) , , &text{if}, , x > 0\
& 0, &text{if} , , x = 0end{cases}$$
Show that $f$ is continuous on $[0,infty)$ and differentiable on $(0,infty)$. Also, show that $f$ has no local maximum or minimum in the endpoint $x = 0$ of the domain of $f$.
I can manage to prove continuity on a single point using the epsilon-delta technique, although the intervals here were a surprise, how do you go about proving such a thing? And any hints about the second part of the problem would also be appreciated.
real-analysis calculus derivatives proof-writing
real-analysis calculus derivatives proof-writing
asked Dec 8 at 9:12
kareem bokai
344
asked Dec 8 at 9:12
kareem bokai
344
edited Dec 8 at 9:27
asked Dec 8 at 9:12
kareem bokai
344
asked Dec 8 at 9:12
kareem bokai
344
asked Dec 8 at 9:12
kareem bokai
344
344
What does the phrase "local maximum or minimum in the endpoint" mean?
– BigbearZzz
Dec 8 at 10:10
According to the definition in the book am using, "A local maximum (minimum) can occur as an end point of the domain of $f$: a point $x$ in the domain of $f$ that does not belong to an open interval lying completely inside the domain". Yeah am equally as confused too :)
– kareem bokai
Dec 8 at 10:19
I don't quite understand what you want to prove here. Perhaps you want to show that $x=0$ is neither a local maximum nor a local minimum?
– BigbearZzz
Dec 8 at 10:21
Yeah, I think the way they formulated the question is horrible, we just have to show that $x = 0$ is neither a local max or min
– kareem bokai
Dec 8 at 10:25
add a comment |
What does the phrase "local maximum or minimum in the endpoint" mean?
– BigbearZzz
Dec 8 at 10:10
According to the definition in the book am using, "A local maximum (minimum) can occur as an end point of the domain of $f$: a point $x$ in the domain of $f$ that does not belong to an open interval lying completely inside the domain". Yeah am equally as confused too :)
– kareem bokai
Dec 8 at 10:19
I don't quite understand what you want to prove here. Perhaps you want to show that $x=0$ is neither a local maximum nor a local minimum?
– BigbearZzz
Dec 8 at 10:21
Yeah, I think the way they formulated the question is horrible, we just have to show that $x = 0$ is neither a local max or min
– kareem bokai
Dec 8 at 10:25
What does the phrase "local maximum or minimum in the endpoint" mean?
– BigbearZzz
Dec 8 at 10:10
What does the phrase "local maximum or minimum in the endpoint" mean?
– BigbearZzz
Dec 8 at 10:10
According to the definition in the book am using, "A local maximum (minimum) can occur as an end point of the domain of $f$: a point $x$ in the domain of $f$ that does not belong to an open interval lying completely inside the domain". Yeah am equally as confused too :)
– kareem bokai
Dec 8 at 10:19
According to the definition in the book am using, "A local maximum (minimum) can occur as an end point of the domain of $f$: a point $x$ in the domain of $f$ that does not belong to an open interval lying completely inside the domain". Yeah am equally as confused too :)
– kareem bokai
Dec 8 at 10:19
I don't quite understand what you want to prove here. Perhaps you want to show that $x=0$ is neither a local maximum nor a local minimum?
– BigbearZzz
Dec 8 at 10:21
I don't quite understand what you want to prove here. Perhaps you want to show that $x=0$ is neither a local maximum nor a local minimum?
– BigbearZzz
Dec 8 at 10:21
Yeah, I think the way they formulated the question is horrible, we just have to show that $x = 0$ is neither a local max or min
– kareem bokai
Dec 8 at 10:25
Yeah, I think the way they formulated the question is horrible, we just have to show that $x = 0$ is neither a local max or min
– kareem bokai
Dec 8 at 10:25
add a comment |
2 Answers
2
active
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votes
An alternative way to show that $f$ is continuous at $x=0$ is to use the sandwich theorem with
$$
-|x| le f(x) le |x|,
$$
where $pm|x|to 0 $ as $xto 0$. Differentiability of $f$ on $(0,infty)$ is obvious.
For the second part, we want to show that for any $varepsilon>0$, the point $x=0$ is neither a maximum nor minimum on $[0,varepsilon)$. Choose $n$ large enough so that $frac 1{2pi n+pi/2}<varepsilon$, then
$$
fleft(frac 1{2pi n+pi/2}right) = frac 1{2pi n+pi/2}>0 = f(0)
$$
so $x=0$ is not a maximum. You should be able to modify this a bit to prove that $x=0$ is not a minimum either. (Hint: we want $sin(1/x)=-1$)
answered Dec 8 at 10:42
BigbearZzz
7,22921648
add a comment |
It'v very easy to show that for any bounded function $g$, defined on a neighbourhood of $0$, the function $f(x)=xcdot g(x)$ is continuous at $x=0$.
For the differentiability of $f$ notice that $(f(x)-f(0))/(x-0)=g(x)$. Now if $lim_{xto0}g(x)$ doesn't exist, $f$ isn't differentiable at $x=0$.
answered Dec 8 at 12:52
Michael Hoppe
10.8k31834
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
An alternative way to show that $f$ is continuous at $x=0$ is to use the sandwich theorem with
$$
-|x| le f(x) le |x|,
$$
where $pm|x|to 0 $ as $xto 0$. Differentiability of $f$ on $(0,infty)$ is obvious.
For the second part, we want to show that for any $varepsilon>0$, the point $x=0$ is neither a maximum nor minimum on $[0,varepsilon)$. Choose $n$ large enough so that $frac 1{2pi n+pi/2}<varepsilon$, then
$$
fleft(frac 1{2pi n+pi/2}right) = frac 1{2pi n+pi/2}>0 = f(0)
$$
so $x=0$ is not a maximum. You should be able to modify this a bit to prove that $x=0$ is not a minimum either. (Hint: we want $sin(1/x)=-1$)
answered Dec 8 at 10:42
BigbearZzz
7,22921648
add a comment |
An alternative way to show that $f$ is continuous at $x=0$ is to use the sandwich theorem with
$$
-|x| le f(x) le |x|,
$$
where $pm|x|to 0 $ as $xto 0$. Differentiability of $f$ on $(0,infty)$ is obvious.
For the second part, we want to show that for any $varepsilon>0$, the point $x=0$ is neither a maximum nor minimum on $[0,varepsilon)$. Choose $n$ large enough so that $frac 1{2pi n+pi/2}<varepsilon$, then
$$
fleft(frac 1{2pi n+pi/2}right) = frac 1{2pi n+pi/2}>0 = f(0)
$$
so $x=0$ is not a maximum. You should be able to modify this a bit to prove that $x=0$ is not a minimum either. (Hint: we want $sin(1/x)=-1$)
answered Dec 8 at 10:42
BigbearZzz
7,22921648
add a comment |
An alternative way to show that $f$ is continuous at $x=0$ is to use the sandwich theorem with
$$
-|x| le f(x) le |x|,
$$
where $pm|x|to 0 $ as $xto 0$. Differentiability of $f$ on $(0,infty)$ is obvious.
For the second part, we want to show that for any $varepsilon>0$, the point $x=0$ is neither a maximum nor minimum on $[0,varepsilon)$. Choose $n$ large enough so that $frac 1{2pi n+pi/2}<varepsilon$, then
$$
fleft(frac 1{2pi n+pi/2}right) = frac 1{2pi n+pi/2}>0 = f(0)
$$
so $x=0$ is not a maximum. You should be able to modify this a bit to prove that $x=0$ is not a minimum either. (Hint: we want $sin(1/x)=-1$)
answered Dec 8 at 10:42
BigbearZzz
7,22921648
An alternative way to show that $f$ is continuous at $x=0$ is to use the sandwich theorem with
$$
-|x| le f(x) le |x|,
$$
where $pm|x|to 0 $ as $xto 0$. Differentiability of $f$ on $(0,infty)$ is obvious.
For the second part, we want to show that for any $varepsilon>0$, the point $x=0$ is neither a maximum nor minimum on $[0,varepsilon)$. Choose $n$ large enough so that $frac 1{2pi n+pi/2}<varepsilon$, then
$$
fleft(frac 1{2pi n+pi/2}right) = frac 1{2pi n+pi/2}>0 = f(0)
$$
so $x=0$ is not a maximum. You should be able to modify this a bit to prove that $x=0$ is not a minimum either. (Hint: we want $sin(1/x)=-1$)
answered Dec 8 at 10:42
BigbearZzz
7,22921648
answered Dec 8 at 10:42
BigbearZzz
7,22921648
answered Dec 8 at 10:42
BigbearZzz
7,22921648
answered Dec 8 at 10:42
BigbearZzz
7,22921648
7,22921648
add a comment |
add a comment |
It'v very easy to show that for any bounded function $g$, defined on a neighbourhood of $0$, the function $f(x)=xcdot g(x)$ is continuous at $x=0$.
For the differentiability of $f$ notice that $(f(x)-f(0))/(x-0)=g(x)$. Now if $lim_{xto0}g(x)$ doesn't exist, $f$ isn't differentiable at $x=0$.
answered Dec 8 at 12:52
Michael Hoppe
10.8k31834
add a comment |
It'v very easy to show that for any bounded function $g$, defined on a neighbourhood of $0$, the function $f(x)=xcdot g(x)$ is continuous at $x=0$.
For the differentiability of $f$ notice that $(f(x)-f(0))/(x-0)=g(x)$. Now if $lim_{xto0}g(x)$ doesn't exist, $f$ isn't differentiable at $x=0$.
answered Dec 8 at 12:52
Michael Hoppe
10.8k31834
add a comment |
It'v very easy to show that for any bounded function $g$, defined on a neighbourhood of $0$, the function $f(x)=xcdot g(x)$ is continuous at $x=0$.
For the differentiability of $f$ notice that $(f(x)-f(0))/(x-0)=g(x)$. Now if $lim_{xto0}g(x)$ doesn't exist, $f$ isn't differentiable at $x=0$.
answered Dec 8 at 12:52
Michael Hoppe
10.8k31834
It'v very easy to show that for any bounded function $g$, defined on a neighbourhood of $0$, the function $f(x)=xcdot g(x)$ is continuous at $x=0$.
For the differentiability of $f$ notice that $(f(x)-f(0))/(x-0)=g(x)$. Now if $lim_{xto0}g(x)$ doesn't exist, $f$ isn't differentiable at $x=0$.
answered Dec 8 at 12:52
Michael Hoppe
10.8k31834
edited Dec 8 at 12:59
answered Dec 8 at 12:52
Michael Hoppe
10.8k31834
answered Dec 8 at 12:52
Michael Hoppe
10.8k31834
answered Dec 8 at 12:52
Michael Hoppe
10.8k31834
10.8k31834
add a comment |
add a comment |
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What does the phrase "local maximum or minimum in the endpoint" mean?
– BigbearZzz
Dec 8 at 10:10
According to the definition in the book am using, "A local maximum (minimum) can occur as an end point of the domain of $f$: a point $x$ in the domain of $f$ that does not belong to an open interval lying completely inside the domain". Yeah am equally as confused too :)
– kareem bokai
Dec 8 at 10:19
I don't quite understand what you want to prove here. Perhaps you want to show that $x=0$ is neither a local maximum nor a local minimum?
– BigbearZzz
Dec 8 at 10:21
Yeah, I think the way they formulated the question is horrible, we just have to show that $x = 0$ is neither a local max or min
– kareem bokai
Dec 8 at 10:25