Mixing
Find range of the given function : $ f(x) = frac{e^x}{1+ lceil xrceil } $ when $ x ge 0 $
1
$begingroup$
Find the range for $ f(x) = cfrac{e^x}{1+lceil x rceil } $ when $xge 0$
My book answers it in a very straight forward manner -
Here $f(x)$ is defined for all $x ge 0$ . Also, $f(x)$ is an increasing function in $[0,infty)$. Thus, range = $[f(0),f(infty)] = [1, infty)$
My question is :
How is this function an increasing function in $[0, infty)$ ?
functions exponential-function
edited Jan 15 at 8:58
Mohammad Zuhair Khan
1,5852625
asked Apr 30 '15 at 3:59
Kushashwa Ravi ShrimaliKushashwa Ravi Shrimali
580315
$endgroup$
add a comment |
1
$begingroup$
Find the range for $ f(x) = cfrac{e^x}{1+lceil x rceil } $ when $xge 0$
My book answers it in a very straight forward manner -
Here $f(x)$ is defined for all $x ge 0$ . Also, $f(x)$ is an increasing function in $[0,infty)$. Thus, range = $[f(0),f(infty)] = [1, infty)$
My question is :
How is this function an increasing function in $[0, infty)$ ?
functions exponential-function
edited Jan 15 at 8:58
Mohammad Zuhair Khan
1,5852625
asked Apr 30 '15 at 3:59
Kushashwa Ravi ShrimaliKushashwa Ravi Shrimali
580315
$endgroup$
$begingroup$
Just to be clear, what do you mean by $[x]$? Presumably the greatest integer $leq x$. Also, why do you think the function is increasing in $[0,infty)$? The question doesn't ask about montonicity.
$endgroup$
– Clayton
Apr 30 '15 at 4:05
$begingroup$
I will edit the question to answer your accordingly. One sec. , Mr. Clayton!
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:06
$begingroup$
I've edited the question, @Clayton . Hope it answers your point.
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:09
1
$begingroup$
It is not increasing, and not continuous (it's easy to prove, but also the plot is quite a good way to see it). You can show it is increasing (strictly) and continuous on every $[n, n+1)$, though, with $f(n+1)<f(n)$. This will be sufficient.
$endgroup$
– Clement C.
Apr 30 '15 at 4:11
add a comment |
1
1
1
$begingroup$
Find the range for $ f(x) = cfrac{e^x}{1+lceil x rceil } $ when $xge 0$
My book answers it in a very straight forward manner -
Here $f(x)$ is defined for all $x ge 0$ . Also, $f(x)$ is an increasing function in $[0,infty)$. Thus, range = $[f(0),f(infty)] = [1, infty)$
My question is :
How is this function an increasing function in $[0, infty)$ ?
functions exponential-function
edited Jan 15 at 8:58
Mohammad Zuhair Khan
1,5852625
asked Apr 30 '15 at 3:59
Kushashwa Ravi ShrimaliKushashwa Ravi Shrimali
580315
$endgroup$
Find the range for $ f(x) = cfrac{e^x}{1+lceil x rceil } $ when $xge 0$
My book answers it in a very straight forward manner -
Here $f(x)$ is defined for all $x ge 0$ . Also, $f(x)$ is an increasing function in $[0,infty)$. Thus, range = $[f(0),f(infty)] = [1, infty)$
My question is :
How is this function an increasing function in $[0, infty)$ ?
functions exponential-function
functions exponential-function
edited Jan 15 at 8:58
Mohammad Zuhair Khan
1,5852625
asked Apr 30 '15 at 3:59
Kushashwa Ravi ShrimaliKushashwa Ravi Shrimali
580315
edited Jan 15 at 8:58
Mohammad Zuhair Khan
1,5852625
asked Apr 30 '15 at 3:59
Kushashwa Ravi ShrimaliKushashwa Ravi Shrimali
580315
edited Jan 15 at 8:58
Mohammad Zuhair Khan
1,5852625
edited Jan 15 at 8:58
Mohammad Zuhair Khan
1,5852625
edited Jan 15 at 8:58
Mohammad Zuhair Khan
1,5852625
1,5852625
asked Apr 30 '15 at 3:59
Kushashwa Ravi ShrimaliKushashwa Ravi Shrimali
580315
asked Apr 30 '15 at 3:59
Kushashwa Ravi ShrimaliKushashwa Ravi Shrimali
580315
asked Apr 30 '15 at 3:59
Kushashwa Ravi ShrimaliKushashwa Ravi Shrimali
580315
580315
$begingroup$
Just to be clear, what do you mean by $[x]$? Presumably the greatest integer $leq x$. Also, why do you think the function is increasing in $[0,infty)$? The question doesn't ask about montonicity.
$endgroup$
– Clayton
Apr 30 '15 at 4:05
$begingroup$
I will edit the question to answer your accordingly. One sec. , Mr. Clayton!
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:06
$begingroup$
I've edited the question, @Clayton . Hope it answers your point.
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:09
1
$begingroup$
It is not increasing, and not continuous (it's easy to prove, but also the plot is quite a good way to see it). You can show it is increasing (strictly) and continuous on every $[n, n+1)$, though, with $f(n+1)<f(n)$. This will be sufficient.
$endgroup$
– Clement C.
Apr 30 '15 at 4:11
add a comment |
$begingroup$
Just to be clear, what do you mean by $[x]$? Presumably the greatest integer $leq x$. Also, why do you think the function is increasing in $[0,infty)$? The question doesn't ask about montonicity.
$endgroup$
– Clayton
Apr 30 '15 at 4:05
$begingroup$
I will edit the question to answer your accordingly. One sec. , Mr. Clayton!
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:06
$begingroup$
I've edited the question, @Clayton . Hope it answers your point.
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:09
1
$begingroup$
It is not increasing, and not continuous (it's easy to prove, but also the plot is quite a good way to see it). You can show it is increasing (strictly) and continuous on every $[n, n+1)$, though, with $f(n+1)<f(n)$. This will be sufficient.
$endgroup$
– Clement C.
Apr 30 '15 at 4:11
$begingroup$
Just to be clear, what do you mean by $[x]$? Presumably the greatest integer $leq x$. Also, why do you think the function is increasing in $[0,infty)$? The question doesn't ask about montonicity.
$endgroup$
– Clayton
Apr 30 '15 at 4:05
$begingroup$
Just to be clear, what do you mean by $[x]$? Presumably the greatest integer $leq x$. Also, why do you think the function is increasing in $[0,infty)$? The question doesn't ask about montonicity.
$endgroup$
– Clayton
Apr 30 '15 at 4:05
$begingroup$
I will edit the question to answer your accordingly. One sec. , Mr. Clayton!
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:06
$begingroup$
I will edit the question to answer your accordingly. One sec. , Mr. Clayton!
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:06
$begingroup$
I've edited the question, @Clayton . Hope it answers your point.
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:09
$begingroup$
I've edited the question, @Clayton . Hope it answers your point.
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:09
1
1
$begingroup$
It is not increasing, and not continuous (it's easy to prove, but also the plot is quite a good way to see it). You can show it is increasing (strictly) and continuous on every $[n, n+1)$, though, with $f(n+1)<f(n)$. This will be sufficient.
$endgroup$
– Clement C.
Apr 30 '15 at 4:11
$begingroup$
It is not increasing, and not continuous (it's easy to prove, but also the plot is quite a good way to see it). You can show it is increasing (strictly) and continuous on every $[n, n+1)$, though, with $f(n+1)<f(n)$. This will be sufficient.
$endgroup$
– Clement C.
Apr 30 '15 at 4:11
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
You have a very good question. For instance, for $x$ just below $1$, we have that
$$ f(0.999) = frac{e^{0.999}}{1} approx e,$$
while for $x = 1$ we have
$$ f(1) = frac{e}{2} < e.$$
So the function is not increasing. But it's not so hard to see that it will always be at least $1$, and it becomes unboundedly large. And at integer values, it temporarily decreases, so that every value in $[1, infty)$ gets hit. I think this is the way to prove the range is $[1, infty)$.
answered Apr 30 '15 at 4:12
davidlowryduda♦davidlowryduda
74.7k7119253
$endgroup$
1
$begingroup$
You will also need some continuity (piecewise) here in order to argue that the full range is reached (that is, that the function is surjective on $[1,infty)$).
$endgroup$
– Clement C.
Apr 30 '15 at 4:15
$begingroup$
@ClementC. Ah, yes. That's very true.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:15
1
$begingroup$
@KushashwaRaviShrimali it will always fail at just-before integer values and the integer values themselves. So $1.999$ compared to $2$, for instance.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:17
1
$begingroup$
Seems like I got it! :) Thanks a lot everyone for your kind help and patience. @ClementC. and mixedmath !
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:33
1
$begingroup$
@KushashwaRaviShrimali: regarding your comment above: every element $x$ of the domain has a unique image $f(x)$ ($f$ couldn't be a function if $x$ were "mapped to more than one $f(x)$"). It is surjective on $[1,infty)$: every element $yin[1,infty)$ has (at least) one preimage $x$ by $f$. It is not injective: some of the elements $yin[1,infty)$ have more than one preimage.
$endgroup$
– Clement C.
Apr 30 '15 at 4:42
|
show 10 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
You have a very good question. For instance, for $x$ just below $1$, we have that
$$ f(0.999) = frac{e^{0.999}}{1} approx e,$$
while for $x = 1$ we have
$$ f(1) = frac{e}{2} < e.$$
So the function is not increasing. But it's not so hard to see that it will always be at least $1$, and it becomes unboundedly large. And at integer values, it temporarily decreases, so that every value in $[1, infty)$ gets hit. I think this is the way to prove the range is $[1, infty)$.
answered Apr 30 '15 at 4:12
davidlowryduda♦davidlowryduda
74.7k7119253
$endgroup$
1
$begingroup$
You will also need some continuity (piecewise) here in order to argue that the full range is reached (that is, that the function is surjective on $[1,infty)$).
$endgroup$
– Clement C.
Apr 30 '15 at 4:15
$begingroup$
@ClementC. Ah, yes. That's very true.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:15
1
$begingroup$
@KushashwaRaviShrimali it will always fail at just-before integer values and the integer values themselves. So $1.999$ compared to $2$, for instance.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:17
1
$begingroup$
Seems like I got it! :) Thanks a lot everyone for your kind help and patience. @ClementC. and mixedmath !
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:33
1
$begingroup$
@KushashwaRaviShrimali: regarding your comment above: every element $x$ of the domain has a unique image $f(x)$ ($f$ couldn't be a function if $x$ were "mapped to more than one $f(x)$"). It is surjective on $[1,infty)$: every element $yin[1,infty)$ has (at least) one preimage $x$ by $f$. It is not injective: some of the elements $yin[1,infty)$ have more than one preimage.
$endgroup$
– Clement C.
Apr 30 '15 at 4:42
|
show 10 more comments
1
$begingroup$
You have a very good question. For instance, for $x$ just below $1$, we have that
$$ f(0.999) = frac{e^{0.999}}{1} approx e,$$
while for $x = 1$ we have
$$ f(1) = frac{e}{2} < e.$$
So the function is not increasing. But it's not so hard to see that it will always be at least $1$, and it becomes unboundedly large. And at integer values, it temporarily decreases, so that every value in $[1, infty)$ gets hit. I think this is the way to prove the range is $[1, infty)$.
answered Apr 30 '15 at 4:12
davidlowryduda♦davidlowryduda
74.7k7119253
$endgroup$
1
$begingroup$
You will also need some continuity (piecewise) here in order to argue that the full range is reached (that is, that the function is surjective on $[1,infty)$).
$endgroup$
– Clement C.
Apr 30 '15 at 4:15
$begingroup$
@ClementC. Ah, yes. That's very true.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:15
1
$begingroup$
@KushashwaRaviShrimali it will always fail at just-before integer values and the integer values themselves. So $1.999$ compared to $2$, for instance.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:17
1
$begingroup$
Seems like I got it! :) Thanks a lot everyone for your kind help and patience. @ClementC. and mixedmath !
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:33
1
$begingroup$
@KushashwaRaviShrimali: regarding your comment above: every element $x$ of the domain has a unique image $f(x)$ ($f$ couldn't be a function if $x$ were "mapped to more than one $f(x)$"). It is surjective on $[1,infty)$: every element $yin[1,infty)$ has (at least) one preimage $x$ by $f$. It is not injective: some of the elements $yin[1,infty)$ have more than one preimage.
$endgroup$
– Clement C.
Apr 30 '15 at 4:42
|
show 10 more comments
1
1
1
$begingroup$
You have a very good question. For instance, for $x$ just below $1$, we have that
$$ f(0.999) = frac{e^{0.999}}{1} approx e,$$
while for $x = 1$ we have
$$ f(1) = frac{e}{2} < e.$$
So the function is not increasing. But it's not so hard to see that it will always be at least $1$, and it becomes unboundedly large. And at integer values, it temporarily decreases, so that every value in $[1, infty)$ gets hit. I think this is the way to prove the range is $[1, infty)$.
answered Apr 30 '15 at 4:12
davidlowryduda♦davidlowryduda
74.7k7119253
$endgroup$
You have a very good question. For instance, for $x$ just below $1$, we have that
$$ f(0.999) = frac{e^{0.999}}{1} approx e,$$
while for $x = 1$ we have
$$ f(1) = frac{e}{2} < e.$$
So the function is not increasing. But it's not so hard to see that it will always be at least $1$, and it becomes unboundedly large. And at integer values, it temporarily decreases, so that every value in $[1, infty)$ gets hit. I think this is the way to prove the range is $[1, infty)$.
answered Apr 30 '15 at 4:12
davidlowryduda♦davidlowryduda
74.7k7119253
edited Apr 30 '15 at 4:16
answered Apr 30 '15 at 4:12
davidlowryduda♦davidlowryduda
74.7k7119253
answered Apr 30 '15 at 4:12
davidlowryduda♦davidlowryduda
74.7k7119253
answered Apr 30 '15 at 4:12
davidlowryduda♦davidlowryduda
74.7k7119253
74.7k7119253
1
$begingroup$
You will also need some continuity (piecewise) here in order to argue that the full range is reached (that is, that the function is surjective on $[1,infty)$).
$endgroup$
– Clement C.
Apr 30 '15 at 4:15
$begingroup$
@ClementC. Ah, yes. That's very true.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:15
1
$begingroup$
@KushashwaRaviShrimali it will always fail at just-before integer values and the integer values themselves. So $1.999$ compared to $2$, for instance.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:17
1
$begingroup$
Seems like I got it! :) Thanks a lot everyone for your kind help and patience. @ClementC. and mixedmath !
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:33
1
$begingroup$
@KushashwaRaviShrimali: regarding your comment above: every element $x$ of the domain has a unique image $f(x)$ ($f$ couldn't be a function if $x$ were "mapped to more than one $f(x)$"). It is surjective on $[1,infty)$: every element $yin[1,infty)$ has (at least) one preimage $x$ by $f$. It is not injective: some of the elements $yin[1,infty)$ have more than one preimage.
$endgroup$
– Clement C.
Apr 30 '15 at 4:42
|
show 10 more comments
1
$begingroup$
You will also need some continuity (piecewise) here in order to argue that the full range is reached (that is, that the function is surjective on $[1,infty)$).
$endgroup$
– Clement C.
Apr 30 '15 at 4:15
$begingroup$
@ClementC. Ah, yes. That's very true.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:15
1
$begingroup$
@KushashwaRaviShrimali it will always fail at just-before integer values and the integer values themselves. So $1.999$ compared to $2$, for instance.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:17
1
$begingroup$
Seems like I got it! :) Thanks a lot everyone for your kind help and patience. @ClementC. and mixedmath !
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:33
1
$begingroup$
@KushashwaRaviShrimali: regarding your comment above: every element $x$ of the domain has a unique image $f(x)$ ($f$ couldn't be a function if $x$ were "mapped to more than one $f(x)$"). It is surjective on $[1,infty)$: every element $yin[1,infty)$ has (at least) one preimage $x$ by $f$. It is not injective: some of the elements $yin[1,infty)$ have more than one preimage.
$endgroup$
– Clement C.
Apr 30 '15 at 4:42
1
1
$begingroup$
You will also need some continuity (piecewise) here in order to argue that the full range is reached (that is, that the function is surjective on $[1,infty)$).
$endgroup$
– Clement C.
Apr 30 '15 at 4:15
$begingroup$
You will also need some continuity (piecewise) here in order to argue that the full range is reached (that is, that the function is surjective on $[1,infty)$).
$endgroup$
– Clement C.
Apr 30 '15 at 4:15
$begingroup$
@ClementC. Ah, yes. That's very true.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:15
$begingroup$
@ClementC. Ah, yes. That's very true.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:15
1
1
$begingroup$
@KushashwaRaviShrimali it will always fail at just-before integer values and the integer values themselves. So $1.999$ compared to $2$, for instance.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:17
$begingroup$
@KushashwaRaviShrimali it will always fail at just-before integer values and the integer values themselves. So $1.999$ compared to $2$, for instance.
$endgroup$
– davidlowryduda♦
Apr 30 '15 at 4:17
1
1
$begingroup$
Seems like I got it! :) Thanks a lot everyone for your kind help and patience. @ClementC. and mixedmath !
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:33
$begingroup$
Seems like I got it! :) Thanks a lot everyone for your kind help and patience. @ClementC. and mixedmath !
$endgroup$
– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:33
1
1
$begingroup$
@KushashwaRaviShrimali: regarding your comment above: every element $x$ of the domain has a unique image $f(x)$ ($f$ couldn't be a function if $x$ were "mapped to more than one $f(x)$"). It is surjective on $[1,infty)$: every element $yin[1,infty)$ has (at least) one preimage $x$ by $f$. It is not injective: some of the elements $yin[1,infty)$ have more than one preimage.
$endgroup$
– Clement C.
Apr 30 '15 at 4:42
$begingroup$
@KushashwaRaviShrimali: regarding your comment above: every element $x$ of the domain has a unique image $f(x)$ ($f$ couldn't be a function if $x$ were "mapped to more than one $f(x)$"). It is surjective on $[1,infty)$: every element $yin[1,infty)$ has (at least) one preimage $x$ by $f$. It is not injective: some of the elements $yin[1,infty)$ have more than one preimage.
$endgroup$
– Clement C.
Apr 30 '15 at 4:42
|
show 10 more comments
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Required, but never shown
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Just to be clear, what do you mean by $[x]$? Presumably the greatest integer $leq x$. Also, why do you think the function is increasing in $[0,infty)$? The question doesn't ask about montonicity.
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– Clayton
Apr 30 '15 at 4:05
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I will edit the question to answer your accordingly. One sec. , Mr. Clayton!
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– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:06
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I've edited the question, @Clayton . Hope it answers your point.
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– Kushashwa Ravi Shrimali
Apr 30 '15 at 4:09
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It is not increasing, and not continuous (it's easy to prove, but also the plot is quite a good way to see it). You can show it is increasing (strictly) and continuous on every $[n, n+1)$, though, with $f(n+1)<f(n)$. This will be sufficient.
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– Clement C.
Apr 30 '15 at 4:11