Mixing
Is $1/x$ a descending function?
$begingroup$
It is descending on all its connected components, but not as a function, right?
But on the other hand the $f'(x) = -x^{-2} < 0$, then it should be descending, what am I missing?
functions
edited Jan 16 at 8:24
Le Anh Dung
1,2071621
asked Jan 16 at 8:15
Jake B.Jake B.
1666
$endgroup$
add a comment |
$begingroup$
It is descending on all its connected components, but not as a function, right?
But on the other hand the $f'(x) = -x^{-2} < 0$, then it should be descending, what am I missing?
functions
edited Jan 16 at 8:24
Le Anh Dung
1,2071621
asked Jan 16 at 8:15
Jake B.Jake B.
1666
$endgroup$
$begingroup$
it is not a function on all of $mathbb{R}$ hence asking for monotonous decreasing only makes sense on its domain, which is $mathbb{R}setminus {0}$
$endgroup$
– Enkidu
Jan 16 at 8:17
add a comment |
$begingroup$
It is descending on all its connected components, but not as a function, right?
But on the other hand the $f'(x) = -x^{-2} < 0$, then it should be descending, what am I missing?
functions
edited Jan 16 at 8:24
Le Anh Dung
1,2071621
asked Jan 16 at 8:15
Jake B.Jake B.
1666
$endgroup$
It is descending on all its connected components, but not as a function, right?
But on the other hand the $f'(x) = -x^{-2} < 0$, then it should be descending, what am I missing?
functions
functions
edited Jan 16 at 8:24
Le Anh Dung
1,2071621
asked Jan 16 at 8:15
Jake B.Jake B.
1666
edited Jan 16 at 8:24
Le Anh Dung
1,2071621
asked Jan 16 at 8:15
Jake B.Jake B.
1666
edited Jan 16 at 8:24
Le Anh Dung
1,2071621
edited Jan 16 at 8:24
Le Anh Dung
1,2071621
edited Jan 16 at 8:24
Le Anh Dung
1,2071621
1,2071621
asked Jan 16 at 8:15
Jake B.Jake B.
1666
asked Jan 16 at 8:15
Jake B.Jake B.
1666
asked Jan 16 at 8:15
Jake B.Jake B.
1666
1666
$begingroup$
it is not a function on all of $mathbb{R}$ hence asking for monotonous decreasing only makes sense on its domain, which is $mathbb{R}setminus {0}$
$endgroup$
– Enkidu
Jan 16 at 8:17
add a comment |
$begingroup$
it is not a function on all of $mathbb{R}$ hence asking for monotonous decreasing only makes sense on its domain, which is $mathbb{R}setminus {0}$
$endgroup$
– Enkidu
Jan 16 at 8:17
$begingroup$
it is not a function on all of $mathbb{R}$ hence asking for monotonous decreasing only makes sense on its domain, which is $mathbb{R}setminus {0}$
$endgroup$
– Enkidu
Jan 16 at 8:17
$begingroup$
it is not a function on all of $mathbb{R}$ hence asking for monotonous decreasing only makes sense on its domain, which is $mathbb{R}setminus {0}$
$endgroup$
– Enkidu
Jan 16 at 8:17
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
The theorem about the relationship between the monotony of $f$ and the sign of $f'$ has to be applied to an interval (a connected set in $mathbb{R}$). Indeed, reading its proof, we have that for $x_1<x_2$, by the Mean Value Theorem, there is some $tin (x_1,x_2)$ such that
$$f(x_2)-f(x_1)=f'(t)(x_2-x_1)$$
So $f(x_2)geq f(x_1)$ if and only if $f'(t)geq 0$.
Here, in order to apply the Mean Value Theorem, we need $f$ to be a continuous function on the closed interval $[x_1,x_2]$ and differentiable on the open interval $(x_1,x_2)$.
edited Jan 22 at 3:42
Pang
13916
answered Jan 16 at 8:17
Robert ZRobert Z
98.5k1068139
$endgroup$
add a comment |
$begingroup$
The theorem which says that $f'(x)<0$ for all $x in I$ implies $f$ is decreasing on $I$ is applicable only when the function is defined and differentiable on that interval. Since your function is not defined at $0$ you cannot apply that theorem to any interval containing $0$. Of course, the function is decreasing in $(0,infty)$ as well as $(-infty,0)$.
answered Jan 16 at 8:18
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
add a comment |
$begingroup$
$frac{1}{x}$ is not defined when $x=0$, so when $x > 0$ it might not be decreasing.
answered Jan 16 at 8:17
Toby MakToby Mak
3,52311128
$endgroup$
add a comment |
$begingroup$
Note that $$f(-1)=-1<1=f(1).$$ Thus $f$ is not decreasing. Is there a contradiction with $f'(x)<0?$ No, since $f$ is not defined at $x=0.$
The domain of $f$ is $(-infty,0)cup (0,infty).$ Since $f'<0$ all we can say is that $f$ is decreasing on $(-infty,0)$ and on $(0,infty).$ But this doesn't imply that $f$ is decreasing on $(-infty,0)cup (0,infty).$
answered Jan 16 at 8:19
mflmfl
26.6k12142
$endgroup$
add a comment |
$begingroup$
graph of 1/x
As you can see from the graph of 1/x , its not defined at 0 but decreasing for other values of x since y'=-(1/x^2)
answered Jan 16 at 8:25
cognitivecognitive
264
$endgroup$
add a comment |
$begingroup$
"Graphical approach"
When you look at graph of $1/x$ you can mark the point for which function not decrease?
answered Jan 16 at 8:24
Kamil KiełczewskiKamil Kiełczewski
1267
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The theorem about the relationship between the monotony of $f$ and the sign of $f'$ has to be applied to an interval (a connected set in $mathbb{R}$). Indeed, reading its proof, we have that for $x_1<x_2$, by the Mean Value Theorem, there is some $tin (x_1,x_2)$ such that
$$f(x_2)-f(x_1)=f'(t)(x_2-x_1)$$
So $f(x_2)geq f(x_1)$ if and only if $f'(t)geq 0$.
Here, in order to apply the Mean Value Theorem, we need $f$ to be a continuous function on the closed interval $[x_1,x_2]$ and differentiable on the open interval $(x_1,x_2)$.
edited Jan 22 at 3:42
Pang
13916
answered Jan 16 at 8:17
Robert ZRobert Z
98.5k1068139
$endgroup$
add a comment |
$begingroup$
The theorem about the relationship between the monotony of $f$ and the sign of $f'$ has to be applied to an interval (a connected set in $mathbb{R}$). Indeed, reading its proof, we have that for $x_1<x_2$, by the Mean Value Theorem, there is some $tin (x_1,x_2)$ such that
$$f(x_2)-f(x_1)=f'(t)(x_2-x_1)$$
So $f(x_2)geq f(x_1)$ if and only if $f'(t)geq 0$.
Here, in order to apply the Mean Value Theorem, we need $f$ to be a continuous function on the closed interval $[x_1,x_2]$ and differentiable on the open interval $(x_1,x_2)$.
edited Jan 22 at 3:42
Pang
13916
answered Jan 16 at 8:17
Robert ZRobert Z
98.5k1068139
$endgroup$
add a comment |
$begingroup$
The theorem about the relationship between the monotony of $f$ and the sign of $f'$ has to be applied to an interval (a connected set in $mathbb{R}$). Indeed, reading its proof, we have that for $x_1<x_2$, by the Mean Value Theorem, there is some $tin (x_1,x_2)$ such that
$$f(x_2)-f(x_1)=f'(t)(x_2-x_1)$$
So $f(x_2)geq f(x_1)$ if and only if $f'(t)geq 0$.
Here, in order to apply the Mean Value Theorem, we need $f$ to be a continuous function on the closed interval $[x_1,x_2]$ and differentiable on the open interval $(x_1,x_2)$.
edited Jan 22 at 3:42
Pang
13916
answered Jan 16 at 8:17
Robert ZRobert Z
98.5k1068139
$endgroup$
The theorem about the relationship between the monotony of $f$ and the sign of $f'$ has to be applied to an interval (a connected set in $mathbb{R}$). Indeed, reading its proof, we have that for $x_1<x_2$, by the Mean Value Theorem, there is some $tin (x_1,x_2)$ such that
$$f(x_2)-f(x_1)=f'(t)(x_2-x_1)$$
So $f(x_2)geq f(x_1)$ if and only if $f'(t)geq 0$.
Here, in order to apply the Mean Value Theorem, we need $f$ to be a continuous function on the closed interval $[x_1,x_2]$ and differentiable on the open interval $(x_1,x_2)$.
edited Jan 22 at 3:42
Pang
13916
answered Jan 16 at 8:17
Robert ZRobert Z
98.5k1068139
edited Jan 22 at 3:42
Pang
13916
edited Jan 22 at 3:42
Pang
13916
edited Jan 22 at 3:42
Pang
13916
13916
answered Jan 16 at 8:17
Robert ZRobert Z
98.5k1068139
answered Jan 16 at 8:17
Robert ZRobert Z
98.5k1068139
answered Jan 16 at 8:17
Robert ZRobert Z
98.5k1068139
98.5k1068139
add a comment |
add a comment |
$begingroup$
The theorem which says that $f'(x)<0$ for all $x in I$ implies $f$ is decreasing on $I$ is applicable only when the function is defined and differentiable on that interval. Since your function is not defined at $0$ you cannot apply that theorem to any interval containing $0$. Of course, the function is decreasing in $(0,infty)$ as well as $(-infty,0)$.
answered Jan 16 at 8:18
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
add a comment |
$begingroup$
The theorem which says that $f'(x)<0$ for all $x in I$ implies $f$ is decreasing on $I$ is applicable only when the function is defined and differentiable on that interval. Since your function is not defined at $0$ you cannot apply that theorem to any interval containing $0$. Of course, the function is decreasing in $(0,infty)$ as well as $(-infty,0)$.
answered Jan 16 at 8:18
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
add a comment |
$begingroup$
The theorem which says that $f'(x)<0$ for all $x in I$ implies $f$ is decreasing on $I$ is applicable only when the function is defined and differentiable on that interval. Since your function is not defined at $0$ you cannot apply that theorem to any interval containing $0$. Of course, the function is decreasing in $(0,infty)$ as well as $(-infty,0)$.
answered Jan 16 at 8:18
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
The theorem which says that $f'(x)<0$ for all $x in I$ implies $f$ is decreasing on $I$ is applicable only when the function is defined and differentiable on that interval. Since your function is not defined at $0$ you cannot apply that theorem to any interval containing $0$. Of course, the function is decreasing in $(0,infty)$ as well as $(-infty,0)$.
answered Jan 16 at 8:18
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
answered Jan 16 at 8:18
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
answered Jan 16 at 8:18
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
answered Jan 16 at 8:18
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
61.7k42262
add a comment |
add a comment |
$begingroup$
$frac{1}{x}$ is not defined when $x=0$, so when $x > 0$ it might not be decreasing.
answered Jan 16 at 8:17
Toby MakToby Mak
3,52311128
$endgroup$
add a comment |
$begingroup$
$frac{1}{x}$ is not defined when $x=0$, so when $x > 0$ it might not be decreasing.
answered Jan 16 at 8:17
Toby MakToby Mak
3,52311128
$endgroup$
add a comment |
$begingroup$
$frac{1}{x}$ is not defined when $x=0$, so when $x > 0$ it might not be decreasing.
answered Jan 16 at 8:17
Toby MakToby Mak
3,52311128
$endgroup$
$frac{1}{x}$ is not defined when $x=0$, so when $x > 0$ it might not be decreasing.
answered Jan 16 at 8:17
Toby MakToby Mak
3,52311128
answered Jan 16 at 8:17
Toby MakToby Mak
3,52311128
answered Jan 16 at 8:17
Toby MakToby Mak
3,52311128
answered Jan 16 at 8:17
Toby MakToby Mak
3,52311128
3,52311128
add a comment |
add a comment |
$begingroup$
Note that $$f(-1)=-1<1=f(1).$$ Thus $f$ is not decreasing. Is there a contradiction with $f'(x)<0?$ No, since $f$ is not defined at $x=0.$
The domain of $f$ is $(-infty,0)cup (0,infty).$ Since $f'<0$ all we can say is that $f$ is decreasing on $(-infty,0)$ and on $(0,infty).$ But this doesn't imply that $f$ is decreasing on $(-infty,0)cup (0,infty).$
answered Jan 16 at 8:19
mflmfl
26.6k12142
$endgroup$
add a comment |
$begingroup$
Note that $$f(-1)=-1<1=f(1).$$ Thus $f$ is not decreasing. Is there a contradiction with $f'(x)<0?$ No, since $f$ is not defined at $x=0.$
The domain of $f$ is $(-infty,0)cup (0,infty).$ Since $f'<0$ all we can say is that $f$ is decreasing on $(-infty,0)$ and on $(0,infty).$ But this doesn't imply that $f$ is decreasing on $(-infty,0)cup (0,infty).$
answered Jan 16 at 8:19
mflmfl
26.6k12142
$endgroup$
add a comment |
$begingroup$
Note that $$f(-1)=-1<1=f(1).$$ Thus $f$ is not decreasing. Is there a contradiction with $f'(x)<0?$ No, since $f$ is not defined at $x=0.$
The domain of $f$ is $(-infty,0)cup (0,infty).$ Since $f'<0$ all we can say is that $f$ is decreasing on $(-infty,0)$ and on $(0,infty).$ But this doesn't imply that $f$ is decreasing on $(-infty,0)cup (0,infty).$
answered Jan 16 at 8:19
mflmfl
26.6k12142
$endgroup$
Note that $$f(-1)=-1<1=f(1).$$ Thus $f$ is not decreasing. Is there a contradiction with $f'(x)<0?$ No, since $f$ is not defined at $x=0.$
The domain of $f$ is $(-infty,0)cup (0,infty).$ Since $f'<0$ all we can say is that $f$ is decreasing on $(-infty,0)$ and on $(0,infty).$ But this doesn't imply that $f$ is decreasing on $(-infty,0)cup (0,infty).$
answered Jan 16 at 8:19
mflmfl
26.6k12142
answered Jan 16 at 8:19
mflmfl
26.6k12142
answered Jan 16 at 8:19
mflmfl
26.6k12142
answered Jan 16 at 8:19
mflmfl
26.6k12142
26.6k12142
add a comment |
add a comment |
$begingroup$
graph of 1/x
As you can see from the graph of 1/x , its not defined at 0 but decreasing for other values of x since y'=-(1/x^2)
answered Jan 16 at 8:25
cognitivecognitive
264
$endgroup$
add a comment |
$begingroup$
graph of 1/x
As you can see from the graph of 1/x , its not defined at 0 but decreasing for other values of x since y'=-(1/x^2)
answered Jan 16 at 8:25
cognitivecognitive
264
$endgroup$
add a comment |
$begingroup$
graph of 1/x
As you can see from the graph of 1/x , its not defined at 0 but decreasing for other values of x since y'=-(1/x^2)
answered Jan 16 at 8:25
cognitivecognitive
264
$endgroup$
graph of 1/x
As you can see from the graph of 1/x , its not defined at 0 but decreasing for other values of x since y'=-(1/x^2)
answered Jan 16 at 8:25
cognitivecognitive
264
answered Jan 16 at 8:25
cognitivecognitive
264
answered Jan 16 at 8:25
cognitivecognitive
264
answered Jan 16 at 8:25
cognitivecognitive
264
264
add a comment |
add a comment |
$begingroup$
"Graphical approach"
When you look at graph of $1/x$ you can mark the point for which function not decrease?
answered Jan 16 at 8:24
Kamil KiełczewskiKamil Kiełczewski
1267
$endgroup$
add a comment |
$begingroup$
"Graphical approach"
When you look at graph of $1/x$ you can mark the point for which function not decrease?
answered Jan 16 at 8:24
Kamil KiełczewskiKamil Kiełczewski
1267
$endgroup$
add a comment |
$begingroup$
"Graphical approach"
When you look at graph of $1/x$ you can mark the point for which function not decrease?
answered Jan 16 at 8:24
Kamil KiełczewskiKamil Kiełczewski
1267
$endgroup$
"Graphical approach"
When you look at graph of $1/x$ you can mark the point for which function not decrease?
answered Jan 16 at 8:24
Kamil KiełczewskiKamil Kiełczewski
1267
answered Jan 16 at 8:24
Kamil KiełczewskiKamil Kiełczewski
1267
answered Jan 16 at 8:24
Kamil KiełczewskiKamil Kiełczewski
1267
answered Jan 16 at 8:24
Kamil KiełczewskiKamil Kiełczewski
1267
1267
add a comment |
add a comment |
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$begingroup$
it is not a function on all of $mathbb{R}$ hence asking for monotonous decreasing only makes sense on its domain, which is $mathbb{R}setminus {0}$
$endgroup$
– Enkidu
Jan 16 at 8:17