Mixing
What is the domain of the function $f(x)=sin^-1left(frac8(3)^x-21-3^2(x-1)right)$?
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
What is the domain of the function
$f(x)=sin^-1left(frac8(3)^x-21-3^2(x-1)right)$?
I started using the fact $-fracpi2le f(x) le fracpi2implies-1 lefrac8(3)^x-21-3^2(x-1)le1$.Now,on dissecting it into two cases.
$CASE (1): -1 lefrac8(3)^x-21-3^2(x-1)$
$CASE (2): frac8(3)^x-21-3^2(x-1)le1$
The calculations is inboth cases are bewidering,that's why i'm not showing it.
I need someone who can help me in solving this.
functions trigonometry inequality
edited 19 hours ago
Shaun
7,30092870
asked 21 hours ago
PK Styles
1,392524
add a comment |Â
up vote
4
down vote
favorite
What is the domain of the function
$f(x)=sin^-1left(frac8(3)^x-21-3^2(x-1)right)$?
I started using the fact $-fracpi2le f(x) le fracpi2implies-1 lefrac8(3)^x-21-3^2(x-1)le1$.Now,on dissecting it into two cases.
$CASE (1): -1 lefrac8(3)^x-21-3^2(x-1)$
$CASE (2): frac8(3)^x-21-3^2(x-1)le1$
The calculations is inboth cases are bewidering,that's why i'm not showing it.
I need someone who can help me in solving this.
functions trigonometry inequality
edited 19 hours ago
Shaun
7,30092870
asked 21 hours ago
PK Styles
1,392524
1
Use$left(fracABright)$for $left(fracABright)$ instead of$(fracAB)$for $(fracAB)$
– Shaun
19 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
What is the domain of the function
$f(x)=sin^-1left(frac8(3)^x-21-3^2(x-1)right)$?
I started using the fact $-fracpi2le f(x) le fracpi2implies-1 lefrac8(3)^x-21-3^2(x-1)le1$.Now,on dissecting it into two cases.
$CASE (1): -1 lefrac8(3)^x-21-3^2(x-1)$
$CASE (2): frac8(3)^x-21-3^2(x-1)le1$
The calculations is inboth cases are bewidering,that's why i'm not showing it.
I need someone who can help me in solving this.
functions trigonometry inequality
edited 19 hours ago
Shaun
7,30092870
asked 21 hours ago
PK Styles
1,392524
What is the domain of the function
$f(x)=sin^-1left(frac8(3)^x-21-3^2(x-1)right)$?
I started using the fact $-fracpi2le f(x) le fracpi2implies-1 lefrac8(3)^x-21-3^2(x-1)le1$.Now,on dissecting it into two cases.
$CASE (1): -1 lefrac8(3)^x-21-3^2(x-1)$
$CASE (2): frac8(3)^x-21-3^2(x-1)le1$
The calculations is inboth cases are bewidering,that's why i'm not showing it.
I need someone who can help me in solving this.
functions trigonometry inequality
edited 19 hours ago
Shaun
7,30092870
asked 21 hours ago
PK Styles
1,392524
edited 19 hours ago
Shaun
7,30092870
edited 19 hours ago
Shaun
7,30092870
edited 19 hours ago
Shaun
7,30092870
7,30092870
asked 21 hours ago
PK Styles
1,392524
asked 21 hours ago
PK Styles
1,392524
asked 21 hours ago
PK Styles
1,392524
1,392524
1
Use$left(fracABright)$for $left(fracABright)$ instead of$(fracAB)$for $(fracAB)$
– Shaun
19 hours ago
add a comment |Â
1
Use$left(fracABright)$for $left(fracABright)$ instead of$(fracAB)$for $(fracAB)$
– Shaun
19 hours ago
1
1
Use
$left(fracABright)$ for $left(fracABright)$ instead of $(fracAB)$ for $(fracAB)$– Shaun
19 hours ago
Use
$left(fracABright)$ for $left(fracABright)$ instead of $(fracAB)$ for $(fracAB)$– Shaun
19 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Hint:
Step1 : Write the first case as $$-1le frac frac 83 cdot t1-t^2$$
Step 2: And second case as $$ frac frac 83 cdot t1-t^2le 1$$
Where $t=3^x-1$ and hence
Step 3: $tge 0$
Now you can take intersection of the intervals obtained from 3 above steps to get domain of $t$ and Resubstitute it in form of $x$ to get original domain
answered 21 hours ago
Manthanein
6,0021336
:Proceeding your work further i get $tin [0,frac13]cup [3,infty)implies 3^x-1 in [0,frac13]cup [3,infty)implies (x-1)in (-infty,-1]cup [1,infty)implies xin (-infty,0]cup [2,infty)$
– PK Styles
19 hours ago
@PKStyles, That is the correct answer.
– prog_SAHIL
17 hours ago
add a comment |Â
up vote
1
down vote
Examine that,
$$lim_xto-inftyfrac8(3)^x-21-3^2(x-1)=0$$
$$lim_xtoinftyfrac8(3)^x-21-3^2(x-1)=0$$
Also the function is discontinuous at $x=1$.
Now check where it assumes the values $1$ and $-1$.
$$frac8(3)^x-21-3^2(x-1)=1$$
$$8(3)^x-2=1-3^2(x-1)$$
$$8(3)^x-2+3^2(x-1)=1$$
$$3^x-1(frac83+3^x-1)=1$$
Let $3^x-1=t$
$$t(8+3t)=3$$
$$t=frac13 implies x=0 $$
$$frac8(3)^x-21-3^2(x-1)=-1$$
$$8(3)^x-2=-1+3^2(x-1)$$
$$8(3)^x-2-3^2(x-1)=-1$$
$$3^x-1(frac83-3^x-1)=-1$$
Let $3^x-1=t$
$$t(8-3t)=-3$$
$$t=3 implies x=2 $$
Therefore, Domain is, $$(-infty,0]cup[2,infty)$$
answered 21 hours ago
prog_SAHIL
732217
add a comment |Â
up vote
0
down vote
Hint:
Let $3^x-2=aimplies a>0$ for real $x$
$$-1ledfrac8a1-9a^2le1$$
$$dfrac8a1-9a^2le1iff0ledfrac9a^2+8a-19a^2-1=dfrac(9a-1)(a+1)(3a-1)(3a+1)$$
As $a.0,$ we need $dfrac9a-13a-1ge0$
$implies$ either $9a-1=0iff a=?$ or $ a>$max$left(dfrac19,dfrac13right)$ or $a<$min$left(dfrac19,dfrac13right)$
$implies x<-2$ or $xge-1$
Similarly for $$-1ledfrac8a1-9a^2iffdfrac8a+1-9a^21-9a^2ge0$$
$$iffdfrac(9a+1)(a-1)(3a+1)(3a-1)le0$$
As $a.0,$ we need $dfraca-13a-1ge0$
Can you take it from here?
answered 21 hours ago
lab bhattacharjee
214k14152263
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint:
Step1 : Write the first case as $$-1le frac frac 83 cdot t1-t^2$$
Step 2: And second case as $$ frac frac 83 cdot t1-t^2le 1$$
Where $t=3^x-1$ and hence
Step 3: $tge 0$
Now you can take intersection of the intervals obtained from 3 above steps to get domain of $t$ and Resubstitute it in form of $x$ to get original domain
answered 21 hours ago
Manthanein
6,0021336
:Proceeding your work further i get $tin [0,frac13]cup [3,infty)implies 3^x-1 in [0,frac13]cup [3,infty)implies (x-1)in (-infty,-1]cup [1,infty)implies xin (-infty,0]cup [2,infty)$
– PK Styles
19 hours ago
@PKStyles, That is the correct answer.
– prog_SAHIL
17 hours ago
add a comment |Â
up vote
2
down vote
accepted
Hint:
Step1 : Write the first case as $$-1le frac frac 83 cdot t1-t^2$$
Step 2: And second case as $$ frac frac 83 cdot t1-t^2le 1$$
Where $t=3^x-1$ and hence
Step 3: $tge 0$
Now you can take intersection of the intervals obtained from 3 above steps to get domain of $t$ and Resubstitute it in form of $x$ to get original domain
answered 21 hours ago
Manthanein
6,0021336
:Proceeding your work further i get $tin [0,frac13]cup [3,infty)implies 3^x-1 in [0,frac13]cup [3,infty)implies (x-1)in (-infty,-1]cup [1,infty)implies xin (-infty,0]cup [2,infty)$
– PK Styles
19 hours ago
@PKStyles, That is the correct answer.
– prog_SAHIL
17 hours ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint:
Step1 : Write the first case as $$-1le frac frac 83 cdot t1-t^2$$
Step 2: And second case as $$ frac frac 83 cdot t1-t^2le 1$$
Where $t=3^x-1$ and hence
Step 3: $tge 0$
Now you can take intersection of the intervals obtained from 3 above steps to get domain of $t$ and Resubstitute it in form of $x$ to get original domain
answered 21 hours ago
Manthanein
6,0021336
Hint:
Step1 : Write the first case as $$-1le frac frac 83 cdot t1-t^2$$
Step 2: And second case as $$ frac frac 83 cdot t1-t^2le 1$$
Where $t=3^x-1$ and hence
Step 3: $tge 0$
Now you can take intersection of the intervals obtained from 3 above steps to get domain of $t$ and Resubstitute it in form of $x$ to get original domain
answered 21 hours ago
Manthanein
6,0021336
answered 21 hours ago
Manthanein
6,0021336
answered 21 hours ago
Manthanein
6,0021336
answered 21 hours ago
Manthanein
6,0021336
6,0021336
:Proceeding your work further i get $tin [0,frac13]cup [3,infty)implies 3^x-1 in [0,frac13]cup [3,infty)implies (x-1)in (-infty,-1]cup [1,infty)implies xin (-infty,0]cup [2,infty)$
– PK Styles
19 hours ago
@PKStyles, That is the correct answer.
– prog_SAHIL
17 hours ago
add a comment |Â
:Proceeding your work further i get $tin [0,frac13]cup [3,infty)implies 3^x-1 in [0,frac13]cup [3,infty)implies (x-1)in (-infty,-1]cup [1,infty)implies xin (-infty,0]cup [2,infty)$
– PK Styles
19 hours ago
@PKStyles, That is the correct answer.
– prog_SAHIL
17 hours ago
:Proceeding your work further i get $tin [0,frac13]cup [3,infty)implies 3^x-1 in [0,frac13]cup [3,infty)implies (x-1)in (-infty,-1]cup [1,infty)implies xin (-infty,0]cup [2,infty)$
– PK Styles
19 hours ago
:Proceeding your work further i get $tin [0,frac13]cup [3,infty)implies 3^x-1 in [0,frac13]cup [3,infty)implies (x-1)in (-infty,-1]cup [1,infty)implies xin (-infty,0]cup [2,infty)$
– PK Styles
19 hours ago
@PKStyles, That is the correct answer.
– prog_SAHIL
17 hours ago
@PKStyles, That is the correct answer.
– prog_SAHIL
17 hours ago
add a comment |Â
up vote
1
down vote
Examine that,
$$lim_xto-inftyfrac8(3)^x-21-3^2(x-1)=0$$
$$lim_xtoinftyfrac8(3)^x-21-3^2(x-1)=0$$
Also the function is discontinuous at $x=1$.
Now check where it assumes the values $1$ and $-1$.
$$frac8(3)^x-21-3^2(x-1)=1$$
$$8(3)^x-2=1-3^2(x-1)$$
$$8(3)^x-2+3^2(x-1)=1$$
$$3^x-1(frac83+3^x-1)=1$$
Let $3^x-1=t$
$$t(8+3t)=3$$
$$t=frac13 implies x=0 $$
$$frac8(3)^x-21-3^2(x-1)=-1$$
$$8(3)^x-2=-1+3^2(x-1)$$
$$8(3)^x-2-3^2(x-1)=-1$$
$$3^x-1(frac83-3^x-1)=-1$$
Let $3^x-1=t$
$$t(8-3t)=-3$$
$$t=3 implies x=2 $$
Therefore, Domain is, $$(-infty,0]cup[2,infty)$$
answered 21 hours ago
prog_SAHIL
732217
add a comment |Â
up vote
1
down vote
Examine that,
$$lim_xto-inftyfrac8(3)^x-21-3^2(x-1)=0$$
$$lim_xtoinftyfrac8(3)^x-21-3^2(x-1)=0$$
Also the function is discontinuous at $x=1$.
Now check where it assumes the values $1$ and $-1$.
$$frac8(3)^x-21-3^2(x-1)=1$$
$$8(3)^x-2=1-3^2(x-1)$$
$$8(3)^x-2+3^2(x-1)=1$$
$$3^x-1(frac83+3^x-1)=1$$
Let $3^x-1=t$
$$t(8+3t)=3$$
$$t=frac13 implies x=0 $$
$$frac8(3)^x-21-3^2(x-1)=-1$$
$$8(3)^x-2=-1+3^2(x-1)$$
$$8(3)^x-2-3^2(x-1)=-1$$
$$3^x-1(frac83-3^x-1)=-1$$
Let $3^x-1=t$
$$t(8-3t)=-3$$
$$t=3 implies x=2 $$
Therefore, Domain is, $$(-infty,0]cup[2,infty)$$
answered 21 hours ago
prog_SAHIL
732217
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Examine that,
$$lim_xto-inftyfrac8(3)^x-21-3^2(x-1)=0$$
$$lim_xtoinftyfrac8(3)^x-21-3^2(x-1)=0$$
Also the function is discontinuous at $x=1$.
Now check where it assumes the values $1$ and $-1$.
$$frac8(3)^x-21-3^2(x-1)=1$$
$$8(3)^x-2=1-3^2(x-1)$$
$$8(3)^x-2+3^2(x-1)=1$$
$$3^x-1(frac83+3^x-1)=1$$
Let $3^x-1=t$
$$t(8+3t)=3$$
$$t=frac13 implies x=0 $$
$$frac8(3)^x-21-3^2(x-1)=-1$$
$$8(3)^x-2=-1+3^2(x-1)$$
$$8(3)^x-2-3^2(x-1)=-1$$
$$3^x-1(frac83-3^x-1)=-1$$
Let $3^x-1=t$
$$t(8-3t)=-3$$
$$t=3 implies x=2 $$
Therefore, Domain is, $$(-infty,0]cup[2,infty)$$
answered 21 hours ago
prog_SAHIL
732217
Examine that,
$$lim_xto-inftyfrac8(3)^x-21-3^2(x-1)=0$$
$$lim_xtoinftyfrac8(3)^x-21-3^2(x-1)=0$$
Also the function is discontinuous at $x=1$.
Now check where it assumes the values $1$ and $-1$.
$$frac8(3)^x-21-3^2(x-1)=1$$
$$8(3)^x-2=1-3^2(x-1)$$
$$8(3)^x-2+3^2(x-1)=1$$
$$3^x-1(frac83+3^x-1)=1$$
Let $3^x-1=t$
$$t(8+3t)=3$$
$$t=frac13 implies x=0 $$
$$frac8(3)^x-21-3^2(x-1)=-1$$
$$8(3)^x-2=-1+3^2(x-1)$$
$$8(3)^x-2-3^2(x-1)=-1$$
$$3^x-1(frac83-3^x-1)=-1$$
Let $3^x-1=t$
$$t(8-3t)=-3$$
$$t=3 implies x=2 $$
Therefore, Domain is, $$(-infty,0]cup[2,infty)$$
answered 21 hours ago
prog_SAHIL
732217
edited 21 hours ago
answered 21 hours ago
prog_SAHIL
732217
answered 21 hours ago
prog_SAHIL
732217
answered 21 hours ago
prog_SAHIL
732217
732217
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint:
Let $3^x-2=aimplies a>0$ for real $x$
$$-1ledfrac8a1-9a^2le1$$
$$dfrac8a1-9a^2le1iff0ledfrac9a^2+8a-19a^2-1=dfrac(9a-1)(a+1)(3a-1)(3a+1)$$
As $a.0,$ we need $dfrac9a-13a-1ge0$
$implies$ either $9a-1=0iff a=?$ or $ a>$max$left(dfrac19,dfrac13right)$ or $a<$min$left(dfrac19,dfrac13right)$
$implies x<-2$ or $xge-1$
Similarly for $$-1ledfrac8a1-9a^2iffdfrac8a+1-9a^21-9a^2ge0$$
$$iffdfrac(9a+1)(a-1)(3a+1)(3a-1)le0$$
As $a.0,$ we need $dfraca-13a-1ge0$
Can you take it from here?
answered 21 hours ago
lab bhattacharjee
214k14152263
add a comment |Â
up vote
0
down vote
Hint:
Let $3^x-2=aimplies a>0$ for real $x$
$$-1ledfrac8a1-9a^2le1$$
$$dfrac8a1-9a^2le1iff0ledfrac9a^2+8a-19a^2-1=dfrac(9a-1)(a+1)(3a-1)(3a+1)$$
As $a.0,$ we need $dfrac9a-13a-1ge0$
$implies$ either $9a-1=0iff a=?$ or $ a>$max$left(dfrac19,dfrac13right)$ or $a<$min$left(dfrac19,dfrac13right)$
$implies x<-2$ or $xge-1$
Similarly for $$-1ledfrac8a1-9a^2iffdfrac8a+1-9a^21-9a^2ge0$$
$$iffdfrac(9a+1)(a-1)(3a+1)(3a-1)le0$$
As $a.0,$ we need $dfraca-13a-1ge0$
Can you take it from here?
answered 21 hours ago
lab bhattacharjee
214k14152263
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
Let $3^x-2=aimplies a>0$ for real $x$
$$-1ledfrac8a1-9a^2le1$$
$$dfrac8a1-9a^2le1iff0ledfrac9a^2+8a-19a^2-1=dfrac(9a-1)(a+1)(3a-1)(3a+1)$$
As $a.0,$ we need $dfrac9a-13a-1ge0$
$implies$ either $9a-1=0iff a=?$ or $ a>$max$left(dfrac19,dfrac13right)$ or $a<$min$left(dfrac19,dfrac13right)$
$implies x<-2$ or $xge-1$
Similarly for $$-1ledfrac8a1-9a^2iffdfrac8a+1-9a^21-9a^2ge0$$
$$iffdfrac(9a+1)(a-1)(3a+1)(3a-1)le0$$
As $a.0,$ we need $dfraca-13a-1ge0$
Can you take it from here?
answered 21 hours ago
lab bhattacharjee
214k14152263
Hint:
Let $3^x-2=aimplies a>0$ for real $x$
$$-1ledfrac8a1-9a^2le1$$
$$dfrac8a1-9a^2le1iff0ledfrac9a^2+8a-19a^2-1=dfrac(9a-1)(a+1)(3a-1)(3a+1)$$
As $a.0,$ we need $dfrac9a-13a-1ge0$
$implies$ either $9a-1=0iff a=?$ or $ a>$max$left(dfrac19,dfrac13right)$ or $a<$min$left(dfrac19,dfrac13right)$
$implies x<-2$ or $xge-1$
Similarly for $$-1ledfrac8a1-9a^2iffdfrac8a+1-9a^21-9a^2ge0$$
$$iffdfrac(9a+1)(a-1)(3a+1)(3a-1)le0$$
As $a.0,$ we need $dfraca-13a-1ge0$
Can you take it from here?
answered 21 hours ago
lab bhattacharjee
214k14152263
answered 21 hours ago
lab bhattacharjee
214k14152263
answered 21 hours ago
lab bhattacharjee
214k14152263
answered 21 hours ago
lab bhattacharjee
214k14152263
214k14152263
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2872879%2fwhat-is-the-domain-of-the-function-fx-sin-1-left-frac83x-21-32%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Use
$left(fracABright)$for $left(fracABright)$ instead of$(fracAB)$for $(fracAB)$– Shaun
19 hours ago