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If the function $f(x)=(x^2+ax+b)lfloor x rfloor$ is continuous in range of (1,4) , then how do I find $a$ and $b$?
How can I find a limit to this sequence? It contains floor functionWhat is the value of $lfloor100Nrfloor$Proving a limit of a trigonometric function: $lim_x to 2/pilfloor sin frac1x rfloor=0$What is the value of $lim _xto 0leftlfloorfractan x sin xx^2rightrfloor$Limit of $f(x)=fracleftlfloor x^2rightrfloor x^2$ functionLimit of Series with differences of Floor functionValue of $left lfloorxright rfloor+left lfloor-xright rfloor$?Prove that f(x) is not continuous and differentiable at any point. Find Riemann sums.How to find: $ limlimits_nrightarrowinfty leftlfloor frac-1nrightrfloor $On continuity of floor, modulus and fractional part function.
$begingroup$
If $$f(x)=(x^2+ax+b)lfloor x rfloor$$
And this function has "CONTINUITY" in this range $$(1,4)$$
So now how to find:
- $a=?$
- $b=?$
Sorrily I don't really know how to do within this hard question. I want to do some with it , but I couldn't. Please help!
limits
edited Mar 16 at 9:56
Bernard
123k741117
asked Mar 16 at 8:26
AquamanAquaman
133
$endgroup$
add a comment |
$begingroup$
If $$f(x)=(x^2+ax+b)lfloor x rfloor$$
And this function has "CONTINUITY" in this range $$(1,4)$$
So now how to find:
- $a=?$
- $b=?$
Sorrily I don't really know how to do within this hard question. I want to do some with it , but I couldn't. Please help!
limits
edited Mar 16 at 9:56
Bernard
123k741117
asked Mar 16 at 8:26
AquamanAquaman
133
$endgroup$
$begingroup$
$lfloor x rfloor$ is discontinuous in $2,3$. To avoid these discontinuities, make null the polynomial in these points.
$endgroup$
– N74
Mar 16 at 11:10
add a comment |
$begingroup$
If $$f(x)=(x^2+ax+b)lfloor x rfloor$$
And this function has "CONTINUITY" in this range $$(1,4)$$
So now how to find:
- $a=?$
- $b=?$
Sorrily I don't really know how to do within this hard question. I want to do some with it , but I couldn't. Please help!
limits
edited Mar 16 at 9:56
Bernard
123k741117
asked Mar 16 at 8:26
AquamanAquaman
133
$endgroup$
If $$f(x)=(x^2+ax+b)lfloor x rfloor$$
And this function has "CONTINUITY" in this range $$(1,4)$$
So now how to find:
- $a=?$
- $b=?$
Sorrily I don't really know how to do within this hard question. I want to do some with it , but I couldn't. Please help!
limits
limits
edited Mar 16 at 9:56
Bernard
123k741117
asked Mar 16 at 8:26
AquamanAquaman
133
edited Mar 16 at 9:56
Bernard
123k741117
asked Mar 16 at 8:26
AquamanAquaman
133
edited Mar 16 at 9:56
Bernard
123k741117
edited Mar 16 at 9:56
Bernard
123k741117
edited Mar 16 at 9:56
Bernard
123k741117
123k741117
asked Mar 16 at 8:26
AquamanAquaman
133
asked Mar 16 at 8:26
AquamanAquaman
133
asked Mar 16 at 8:26
AquamanAquaman
133
133
$begingroup$
$lfloor x rfloor$ is discontinuous in $2,3$. To avoid these discontinuities, make null the polynomial in these points.
$endgroup$
– N74
Mar 16 at 11:10
add a comment |
$begingroup$
$lfloor x rfloor$ is discontinuous in $2,3$. To avoid these discontinuities, make null the polynomial in these points.
$endgroup$
– N74
Mar 16 at 11:10
$begingroup$
$lfloor x rfloor$ is discontinuous in $2,3$. To avoid these discontinuities, make null the polynomial in these points.
$endgroup$
– N74
Mar 16 at 11:10
$begingroup$
$lfloor x rfloor$ is discontinuous in $2,3$. To avoid these discontinuities, make null the polynomial in these points.
$endgroup$
– N74
Mar 16 at 11:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The point is that continuity anyway holds at non-integer points, because here $lfloor xrfloor$ is continuous(why?) and $x^2+ax+b$ is continuous anyway, so $f$ being the product of these will be continuous.
The only problematic points will then be $2$ and $3$ which lie in the domain. For example, for continuity at $2$ you just need to find what $a$ and $b$ satisfy $lim_x to 2^+ f(x) = lim_x to 2^- f(x) = f(2)$.
But note that $lim_x to 2^+ f= 2(2^2+2a+b)$ and $lim_x to 2^- f = 3(2^2+2a+b)$. Therefore, you get $4+2a+b = 0$ by equating these.
Similarly, doing this for $3$ gives $9+3a+b = 0$. Now you have two simultaneous equations in two variables.
answered Mar 16 at 8:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
add a comment |
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$begingroup$
The point is that continuity anyway holds at non-integer points, because here $lfloor xrfloor$ is continuous(why?) and $x^2+ax+b$ is continuous anyway, so $f$ being the product of these will be continuous.
The only problematic points will then be $2$ and $3$ which lie in the domain. For example, for continuity at $2$ you just need to find what $a$ and $b$ satisfy $lim_x to 2^+ f(x) = lim_x to 2^- f(x) = f(2)$.
But note that $lim_x to 2^+ f= 2(2^2+2a+b)$ and $lim_x to 2^- f = 3(2^2+2a+b)$. Therefore, you get $4+2a+b = 0$ by equating these.
Similarly, doing this for $3$ gives $9+3a+b = 0$. Now you have two simultaneous equations in two variables.
answered Mar 16 at 8:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
add a comment |
$begingroup$
The point is that continuity anyway holds at non-integer points, because here $lfloor xrfloor$ is continuous(why?) and $x^2+ax+b$ is continuous anyway, so $f$ being the product of these will be continuous.
The only problematic points will then be $2$ and $3$ which lie in the domain. For example, for continuity at $2$ you just need to find what $a$ and $b$ satisfy $lim_x to 2^+ f(x) = lim_x to 2^- f(x) = f(2)$.
But note that $lim_x to 2^+ f= 2(2^2+2a+b)$ and $lim_x to 2^- f = 3(2^2+2a+b)$. Therefore, you get $4+2a+b = 0$ by equating these.
Similarly, doing this for $3$ gives $9+3a+b = 0$. Now you have two simultaneous equations in two variables.
answered Mar 16 at 8:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
add a comment |
$begingroup$
The point is that continuity anyway holds at non-integer points, because here $lfloor xrfloor$ is continuous(why?) and $x^2+ax+b$ is continuous anyway, so $f$ being the product of these will be continuous.
The only problematic points will then be $2$ and $3$ which lie in the domain. For example, for continuity at $2$ you just need to find what $a$ and $b$ satisfy $lim_x to 2^+ f(x) = lim_x to 2^- f(x) = f(2)$.
But note that $lim_x to 2^+ f= 2(2^2+2a+b)$ and $lim_x to 2^- f = 3(2^2+2a+b)$. Therefore, you get $4+2a+b = 0$ by equating these.
Similarly, doing this for $3$ gives $9+3a+b = 0$. Now you have two simultaneous equations in two variables.
answered Mar 16 at 8:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
The point is that continuity anyway holds at non-integer points, because here $lfloor xrfloor$ is continuous(why?) and $x^2+ax+b$ is continuous anyway, so $f$ being the product of these will be continuous.
The only problematic points will then be $2$ and $3$ which lie in the domain. For example, for continuity at $2$ you just need to find what $a$ and $b$ satisfy $lim_x to 2^+ f(x) = lim_x to 2^- f(x) = f(2)$.
But note that $lim_x to 2^+ f= 2(2^2+2a+b)$ and $lim_x to 2^- f = 3(2^2+2a+b)$. Therefore, you get $4+2a+b = 0$ by equating these.
Similarly, doing this for $3$ gives $9+3a+b = 0$. Now you have two simultaneous equations in two variables.
answered Mar 16 at 8:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
answered Mar 16 at 8:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
answered Mar 16 at 8:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
answered Mar 16 at 8:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
39.8k33477
add a comment |
add a comment |
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$begingroup$
$lfloor x rfloor$ is discontinuous in $2,3$. To avoid these discontinuities, make null the polynomial in these points.
$endgroup$
– N74
Mar 16 at 11:10