Number of discontinuities of $lfloorsin ^{-1} xrfloor+lfloorcos ^{-1} xrfloor$
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Find Number of discontinuities of $$f(x)=lfloorsin ^{-1} xrfloor+lfloorcos ^{-1} xrfloor$$
My try:
The floor function $lfloor xrfloor$ is discontinuous when $x in mathbb{Z}$
Now the integer values $sin^{-1}x$ takes are ${-1,0,1}$
This means the function $h(x)=lfloor sin^{-1} xrfloor$ is discontinuous at
$x=-sin 1 , 0, sin 1$
Similarly the integer values $cos^{-1} x$ takes are $0,1,2,3$
This means the function $g(x)=lfloor cos^{-1} xrfloor$ is discontinuos at $x=cos 3, cos 2,cos 1,1$
but since the domain of $f(x)$ is $[-1, :: 1]$
we need to check Left continuity of $f(x)$ at $x=1$
$$lim _{x to 1^-}f(x)=lim_{h to 0} lfloor sin^{-1}(1-h)rfloor+lfloor cos^{-1}(1-h)rfloor=1=f(1)$$
hence $f(x)$ is continuous at $x=1$
hence number of discontinuities are $6$
is this right approach?
algebra-precalculus limits derivatives continuity
asked Dec 5 at 17:11
Umesh shankar
2,50111219
add a comment |
up vote
3
down vote
favorite
Find Number of discontinuities of $$f(x)=lfloorsin ^{-1} xrfloor+lfloorcos ^{-1} xrfloor$$
My try:
The floor function $lfloor xrfloor$ is discontinuous when $x in mathbb{Z}$
Now the integer values $sin^{-1}x$ takes are ${-1,0,1}$
This means the function $h(x)=lfloor sin^{-1} xrfloor$ is discontinuous at
$x=-sin 1 , 0, sin 1$
Similarly the integer values $cos^{-1} x$ takes are $0,1,2,3$
This means the function $g(x)=lfloor cos^{-1} xrfloor$ is discontinuos at $x=cos 3, cos 2,cos 1,1$
but since the domain of $f(x)$ is $[-1, :: 1]$
we need to check Left continuity of $f(x)$ at $x=1$
$$lim _{x to 1^-}f(x)=lim_{h to 0} lfloor sin^{-1}(1-h)rfloor+lfloor cos^{-1}(1-h)rfloor=1=f(1)$$
hence $f(x)$ is continuous at $x=1$
hence number of discontinuities are $6$
is this right approach?
algebra-precalculus limits derivatives continuity
asked Dec 5 at 17:11
Umesh shankar
2,50111219
2
There's no indication of why the function shouldn't be continuous at those points. You only found the points where the discontinuity can happen, but you need to check if the function really is discontinuous at those points.
– Jakobian
Dec 5 at 17:34
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Find Number of discontinuities of $$f(x)=lfloorsin ^{-1} xrfloor+lfloorcos ^{-1} xrfloor$$
My try:
The floor function $lfloor xrfloor$ is discontinuous when $x in mathbb{Z}$
Now the integer values $sin^{-1}x$ takes are ${-1,0,1}$
This means the function $h(x)=lfloor sin^{-1} xrfloor$ is discontinuous at
$x=-sin 1 , 0, sin 1$
Similarly the integer values $cos^{-1} x$ takes are $0,1,2,3$
This means the function $g(x)=lfloor cos^{-1} xrfloor$ is discontinuos at $x=cos 3, cos 2,cos 1,1$
but since the domain of $f(x)$ is $[-1, :: 1]$
we need to check Left continuity of $f(x)$ at $x=1$
$$lim _{x to 1^-}f(x)=lim_{h to 0} lfloor sin^{-1}(1-h)rfloor+lfloor cos^{-1}(1-h)rfloor=1=f(1)$$
hence $f(x)$ is continuous at $x=1$
hence number of discontinuities are $6$
is this right approach?
algebra-precalculus limits derivatives continuity
asked Dec 5 at 17:11
Umesh shankar
2,50111219
Find Number of discontinuities of $$f(x)=lfloorsin ^{-1} xrfloor+lfloorcos ^{-1} xrfloor$$
My try:
The floor function $lfloor xrfloor$ is discontinuous when $x in mathbb{Z}$
Now the integer values $sin^{-1}x$ takes are ${-1,0,1}$
This means the function $h(x)=lfloor sin^{-1} xrfloor$ is discontinuous at
$x=-sin 1 , 0, sin 1$
Similarly the integer values $cos^{-1} x$ takes are $0,1,2,3$
This means the function $g(x)=lfloor cos^{-1} xrfloor$ is discontinuos at $x=cos 3, cos 2,cos 1,1$
but since the domain of $f(x)$ is $[-1, :: 1]$
we need to check Left continuity of $f(x)$ at $x=1$
$$lim _{x to 1^-}f(x)=lim_{h to 0} lfloor sin^{-1}(1-h)rfloor+lfloor cos^{-1}(1-h)rfloor=1=f(1)$$
hence $f(x)$ is continuous at $x=1$
hence number of discontinuities are $6$
is this right approach?
algebra-precalculus limits derivatives continuity
algebra-precalculus limits derivatives continuity
asked Dec 5 at 17:11
Umesh shankar
2,50111219
asked Dec 5 at 17:11
Umesh shankar
2,50111219
asked Dec 5 at 17:11
Umesh shankar
2,50111219
asked Dec 5 at 17:11
Umesh shankar
2,50111219
asked Dec 5 at 17:11
Umesh shankar
2,50111219
2,50111219
2
There's no indication of why the function shouldn't be continuous at those points. You only found the points where the discontinuity can happen, but you need to check if the function really is discontinuous at those points.
– Jakobian
Dec 5 at 17:34
add a comment |
2
There's no indication of why the function shouldn't be continuous at those points. You only found the points where the discontinuity can happen, but you need to check if the function really is discontinuous at those points.
– Jakobian
Dec 5 at 17:34
2
2
There's no indication of why the function shouldn't be continuous at those points. You only found the points where the discontinuity can happen, but you need to check if the function really is discontinuous at those points.
– Jakobian
Dec 5 at 17:34
There's no indication of why the function shouldn't be continuous at those points. You only found the points where the discontinuity can happen, but you need to check if the function really is discontinuous at those points.
– Jakobian
Dec 5 at 17:34
add a comment |
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2
There's no indication of why the function shouldn't be continuous at those points. You only found the points where the discontinuity can happen, but you need to check if the function really is discontinuous at those points.
– Jakobian
Dec 5 at 17:34