Find the image and preimage of some functions
$begingroup$
I have functions $f: mathbb{R} rightarrow mathbb{R}; f(x) = x^2-4x-1$ and $g:mathbb{C} rightarrow mathbb{C}; g(x) = x^5$ and the set $D={z in mathbb{C} | |z|<1}$.
I need to find $f((1,infty))$, $f^{-1}((1,infty))$ and $g(D)$. My instructor taught me a method in which I need to intuitivly find the image/preimage and then to show formally that my intuition is right by double-inclusion. For example I say that $f((1,infty))=[-4,infty)$, and then I would show this by double-inclusion. I find this method very confusing. Is there any method and if not could you please give me some indications. Thank you.
functions
asked Dec 6 '16 at 12:45
Raducu MihaiRaducu Mihai
345210
$endgroup$
add a comment |
$begingroup$
I have functions $f: mathbb{R} rightarrow mathbb{R}; f(x) = x^2-4x-1$ and $g:mathbb{C} rightarrow mathbb{C}; g(x) = x^5$ and the set $D={z in mathbb{C} | |z|<1}$.
I need to find $f((1,infty))$, $f^{-1}((1,infty))$ and $g(D)$. My instructor taught me a method in which I need to intuitivly find the image/preimage and then to show formally that my intuition is right by double-inclusion. For example I say that $f((1,infty))=[-4,infty)$, and then I would show this by double-inclusion. I find this method very confusing. Is there any method and if not could you please give me some indications. Thank you.
functions
asked Dec 6 '16 at 12:45
Raducu MihaiRaducu Mihai
345210
$endgroup$
$begingroup$
Does $f^{-1}$ mean inverse of $f$?
$endgroup$
– Arnaldo
Dec 6 '16 at 12:54
$begingroup$
No, its the preimage .
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:07
add a comment |
$begingroup$
I have functions $f: mathbb{R} rightarrow mathbb{R}; f(x) = x^2-4x-1$ and $g:mathbb{C} rightarrow mathbb{C}; g(x) = x^5$ and the set $D={z in mathbb{C} | |z|<1}$.
I need to find $f((1,infty))$, $f^{-1}((1,infty))$ and $g(D)$. My instructor taught me a method in which I need to intuitivly find the image/preimage and then to show formally that my intuition is right by double-inclusion. For example I say that $f((1,infty))=[-4,infty)$, and then I would show this by double-inclusion. I find this method very confusing. Is there any method and if not could you please give me some indications. Thank you.
functions
asked Dec 6 '16 at 12:45
Raducu MihaiRaducu Mihai
345210
$endgroup$
I have functions $f: mathbb{R} rightarrow mathbb{R}; f(x) = x^2-4x-1$ and $g:mathbb{C} rightarrow mathbb{C}; g(x) = x^5$ and the set $D={z in mathbb{C} | |z|<1}$.
I need to find $f((1,infty))$, $f^{-1}((1,infty))$ and $g(D)$. My instructor taught me a method in which I need to intuitivly find the image/preimage and then to show formally that my intuition is right by double-inclusion. For example I say that $f((1,infty))=[-4,infty)$, and then I would show this by double-inclusion. I find this method very confusing. Is there any method and if not could you please give me some indications. Thank you.
functions
functions
asked Dec 6 '16 at 12:45
Raducu MihaiRaducu Mihai
345210
asked Dec 6 '16 at 12:45
Raducu MihaiRaducu Mihai
345210
asked Dec 6 '16 at 12:45
Raducu MihaiRaducu Mihai
345210
asked Dec 6 '16 at 12:45
Raducu MihaiRaducu Mihai
345210
asked Dec 6 '16 at 12:45
Raducu MihaiRaducu Mihai
345210
345210
$begingroup$
Does $f^{-1}$ mean inverse of $f$?
$endgroup$
– Arnaldo
Dec 6 '16 at 12:54
$begingroup$
No, its the preimage .
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:07
add a comment |
$begingroup$
Does $f^{-1}$ mean inverse of $f$?
$endgroup$
– Arnaldo
Dec 6 '16 at 12:54
$begingroup$
No, its the preimage .
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:07
$begingroup$
Does $f^{-1}$ mean inverse of $f$?
$endgroup$
– Arnaldo
Dec 6 '16 at 12:54
$begingroup$
Does $f^{-1}$ mean inverse of $f$?
$endgroup$
– Arnaldo
Dec 6 '16 at 12:54
$begingroup$
No, its the preimage .
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:07
$begingroup$
No, its the preimage .
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I agree with your instructor. His approach is reasonable here. Now a few hints.
1) Draw a graph of the parabola $f(x) = x^2-4x-1=(x-2)^2-5$.
What are your candidates for
$f((1,infty))$ and $f^{-1}((1,infty))$?
2) As regards $g(z)=z^5$ try to show that $g(D)=D$.
$g(D)subseteq D$ because if $|z|<1$ then $|g(z)|=|z^5|=|z|^5<1$.
For the other inclusion, we have that
if $re^{it}in D$ then $0leq r<1$ and $tin [0,2pi)$.
Can you find $zin D$ such that $z^5=re^{it}$?
answered Dec 6 '16 at 12:50
Robert ZRobert Z
96.7k1066137
$endgroup$
$begingroup$
I graphed the function using this site desmos.com/calculator . And it looks like $f((1,infty)) = [-4,infty)$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:18
$begingroup$
@Raducu Mihai What is the value of $f(2)$?
$endgroup$
– Robert Z
Dec 6 '16 at 13:22
$begingroup$
The value of $f(2) = -5$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:24
$begingroup$
Now $2in(1,infty)$ but $-5not in [-4,+infty)$. Something is wrong in your answer.
$endgroup$
– Robert Z
Dec 6 '16 at 13:25
$begingroup$
Oook I got it. The interval is open at 1. Yeah I see now
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:25
|
show 4 more comments
$begingroup$
I also agree with your instructor.
$1)$ The graph of $f$ is a parabola with minimal point at $f(2)=-5$ and up concavity. See that $f(1)=-4$:
$A=(1,0)$ give us the point $B=(1,f(1)=-4)$ and then is easy to see that $f((1,infty))=[-5,infty[$. For $f^{-1}((1,infty))$ we have to see what are the values of $x$ such that $f(x) in (1,infty)$. The point $F$ represent the initial point. It is the point such that
$f(x)=1 Rightarrow x^2-4x-1=1 Rightarrow x=2+sqrt{6}$ and $x=2-sqrt{6}$
That means if we take $x<2-sqrt{6}$ or $x>2+sqrt{6}$ we will get $f(x) in (1,infty)$.
$2)$ If $|z|<1$ write $z=r(cosalpha+isinalpha)$, with $r<1$ and $alpha in [0,2pi]$ and then $g(z)=z^5=r^5(cos5alpha+isin5alpha))$ with, $r^5<1$ and $5alpha in [0,5pi]$.
answered Dec 6 '16 at 14:07
ArnaldoArnaldo
18.2k42246
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
$begingroup$
I agree with your instructor. His approach is reasonable here. Now a few hints.
1) Draw a graph of the parabola $f(x) = x^2-4x-1=(x-2)^2-5$.
What are your candidates for
$f((1,infty))$ and $f^{-1}((1,infty))$?
2) As regards $g(z)=z^5$ try to show that $g(D)=D$.
$g(D)subseteq D$ because if $|z|<1$ then $|g(z)|=|z^5|=|z|^5<1$.
For the other inclusion, we have that
if $re^{it}in D$ then $0leq r<1$ and $tin [0,2pi)$.
Can you find $zin D$ such that $z^5=re^{it}$?
answered Dec 6 '16 at 12:50
Robert ZRobert Z
96.7k1066137
$endgroup$
$begingroup$
I graphed the function using this site desmos.com/calculator . And it looks like $f((1,infty)) = [-4,infty)$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:18
$begingroup$
@Raducu Mihai What is the value of $f(2)$?
$endgroup$
– Robert Z
Dec 6 '16 at 13:22
$begingroup$
The value of $f(2) = -5$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:24
$begingroup$
Now $2in(1,infty)$ but $-5not in [-4,+infty)$. Something is wrong in your answer.
$endgroup$
– Robert Z
Dec 6 '16 at 13:25
$begingroup$
Oook I got it. The interval is open at 1. Yeah I see now
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:25
|
show 4 more comments
$begingroup$
I agree with your instructor. His approach is reasonable here. Now a few hints.
1) Draw a graph of the parabola $f(x) = x^2-4x-1=(x-2)^2-5$.
What are your candidates for
$f((1,infty))$ and $f^{-1}((1,infty))$?
2) As regards $g(z)=z^5$ try to show that $g(D)=D$.
$g(D)subseteq D$ because if $|z|<1$ then $|g(z)|=|z^5|=|z|^5<1$.
For the other inclusion, we have that
if $re^{it}in D$ then $0leq r<1$ and $tin [0,2pi)$.
Can you find $zin D$ such that $z^5=re^{it}$?
answered Dec 6 '16 at 12:50
Robert ZRobert Z
96.7k1066137
$endgroup$
$begingroup$
I graphed the function using this site desmos.com/calculator . And it looks like $f((1,infty)) = [-4,infty)$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:18
$begingroup$
@Raducu Mihai What is the value of $f(2)$?
$endgroup$
– Robert Z
Dec 6 '16 at 13:22
$begingroup$
The value of $f(2) = -5$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:24
$begingroup$
Now $2in(1,infty)$ but $-5not in [-4,+infty)$. Something is wrong in your answer.
$endgroup$
– Robert Z
Dec 6 '16 at 13:25
$begingroup$
Oook I got it. The interval is open at 1. Yeah I see now
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:25
|
show 4 more comments
$begingroup$
I agree with your instructor. His approach is reasonable here. Now a few hints.
1) Draw a graph of the parabola $f(x) = x^2-4x-1=(x-2)^2-5$.
What are your candidates for
$f((1,infty))$ and $f^{-1}((1,infty))$?
2) As regards $g(z)=z^5$ try to show that $g(D)=D$.
$g(D)subseteq D$ because if $|z|<1$ then $|g(z)|=|z^5|=|z|^5<1$.
For the other inclusion, we have that
if $re^{it}in D$ then $0leq r<1$ and $tin [0,2pi)$.
Can you find $zin D$ such that $z^5=re^{it}$?
answered Dec 6 '16 at 12:50
Robert ZRobert Z
96.7k1066137
$endgroup$
I agree with your instructor. His approach is reasonable here. Now a few hints.
1) Draw a graph of the parabola $f(x) = x^2-4x-1=(x-2)^2-5$.
What are your candidates for
$f((1,infty))$ and $f^{-1}((1,infty))$?
2) As regards $g(z)=z^5$ try to show that $g(D)=D$.
$g(D)subseteq D$ because if $|z|<1$ then $|g(z)|=|z^5|=|z|^5<1$.
For the other inclusion, we have that
if $re^{it}in D$ then $0leq r<1$ and $tin [0,2pi)$.
Can you find $zin D$ such that $z^5=re^{it}$?
answered Dec 6 '16 at 12:50
Robert ZRobert Z
96.7k1066137
edited Dec 6 '16 at 14:00
answered Dec 6 '16 at 12:50
Robert ZRobert Z
96.7k1066137
answered Dec 6 '16 at 12:50
Robert ZRobert Z
96.7k1066137
answered Dec 6 '16 at 12:50
Robert ZRobert Z
96.7k1066137
96.7k1066137
$begingroup$
I graphed the function using this site desmos.com/calculator . And it looks like $f((1,infty)) = [-4,infty)$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:18
$begingroup$
@Raducu Mihai What is the value of $f(2)$?
$endgroup$
– Robert Z
Dec 6 '16 at 13:22
$begingroup$
The value of $f(2) = -5$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:24
$begingroup$
Now $2in(1,infty)$ but $-5not in [-4,+infty)$. Something is wrong in your answer.
$endgroup$
– Robert Z
Dec 6 '16 at 13:25
$begingroup$
Oook I got it. The interval is open at 1. Yeah I see now
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:25
|
show 4 more comments
$begingroup$
I graphed the function using this site desmos.com/calculator . And it looks like $f((1,infty)) = [-4,infty)$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:18
$begingroup$
@Raducu Mihai What is the value of $f(2)$?
$endgroup$
– Robert Z
Dec 6 '16 at 13:22
$begingroup$
The value of $f(2) = -5$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:24
$begingroup$
Now $2in(1,infty)$ but $-5not in [-4,+infty)$. Something is wrong in your answer.
$endgroup$
– Robert Z
Dec 6 '16 at 13:25
$begingroup$
Oook I got it. The interval is open at 1. Yeah I see now
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:25
$begingroup$
I graphed the function using this site desmos.com/calculator . And it looks like $f((1,infty)) = [-4,infty)$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:18
$begingroup$
I graphed the function using this site desmos.com/calculator . And it looks like $f((1,infty)) = [-4,infty)$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:18
$begingroup$
@Raducu Mihai What is the value of $f(2)$?
$endgroup$
– Robert Z
Dec 6 '16 at 13:22
$begingroup$
@Raducu Mihai What is the value of $f(2)$?
$endgroup$
– Robert Z
Dec 6 '16 at 13:22
$begingroup$
The value of $f(2) = -5$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:24
$begingroup$
The value of $f(2) = -5$
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:24
$begingroup$
Now $2in(1,infty)$ but $-5not in [-4,+infty)$. Something is wrong in your answer.
$endgroup$
– Robert Z
Dec 6 '16 at 13:25
$begingroup$
Now $2in(1,infty)$ but $-5not in [-4,+infty)$. Something is wrong in your answer.
$endgroup$
– Robert Z
Dec 6 '16 at 13:25
$begingroup$
Oook I got it. The interval is open at 1. Yeah I see now
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:25
$begingroup$
Oook I got it. The interval is open at 1. Yeah I see now
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:25
|
show 4 more comments
$begingroup$
I also agree with your instructor.
$1)$ The graph of $f$ is a parabola with minimal point at $f(2)=-5$ and up concavity. See that $f(1)=-4$:
$A=(1,0)$ give us the point $B=(1,f(1)=-4)$ and then is easy to see that $f((1,infty))=[-5,infty[$. For $f^{-1}((1,infty))$ we have to see what are the values of $x$ such that $f(x) in (1,infty)$. The point $F$ represent the initial point. It is the point such that
$f(x)=1 Rightarrow x^2-4x-1=1 Rightarrow x=2+sqrt{6}$ and $x=2-sqrt{6}$
That means if we take $x<2-sqrt{6}$ or $x>2+sqrt{6}$ we will get $f(x) in (1,infty)$.
$2)$ If $|z|<1$ write $z=r(cosalpha+isinalpha)$, with $r<1$ and $alpha in [0,2pi]$ and then $g(z)=z^5=r^5(cos5alpha+isin5alpha))$ with, $r^5<1$ and $5alpha in [0,5pi]$.
answered Dec 6 '16 at 14:07
ArnaldoArnaldo
18.2k42246
$endgroup$
add a comment |
$begingroup$
I also agree with your instructor.
$1)$ The graph of $f$ is a parabola with minimal point at $f(2)=-5$ and up concavity. See that $f(1)=-4$:
$A=(1,0)$ give us the point $B=(1,f(1)=-4)$ and then is easy to see that $f((1,infty))=[-5,infty[$. For $f^{-1}((1,infty))$ we have to see what are the values of $x$ such that $f(x) in (1,infty)$. The point $F$ represent the initial point. It is the point such that
$f(x)=1 Rightarrow x^2-4x-1=1 Rightarrow x=2+sqrt{6}$ and $x=2-sqrt{6}$
That means if we take $x<2-sqrt{6}$ or $x>2+sqrt{6}$ we will get $f(x) in (1,infty)$.
$2)$ If $|z|<1$ write $z=r(cosalpha+isinalpha)$, with $r<1$ and $alpha in [0,2pi]$ and then $g(z)=z^5=r^5(cos5alpha+isin5alpha))$ with, $r^5<1$ and $5alpha in [0,5pi]$.
answered Dec 6 '16 at 14:07
ArnaldoArnaldo
18.2k42246
$endgroup$
add a comment |
$begingroup$
I also agree with your instructor.
$1)$ The graph of $f$ is a parabola with minimal point at $f(2)=-5$ and up concavity. See that $f(1)=-4$:
$A=(1,0)$ give us the point $B=(1,f(1)=-4)$ and then is easy to see that $f((1,infty))=[-5,infty[$. For $f^{-1}((1,infty))$ we have to see what are the values of $x$ such that $f(x) in (1,infty)$. The point $F$ represent the initial point. It is the point such that
$f(x)=1 Rightarrow x^2-4x-1=1 Rightarrow x=2+sqrt{6}$ and $x=2-sqrt{6}$
That means if we take $x<2-sqrt{6}$ or $x>2+sqrt{6}$ we will get $f(x) in (1,infty)$.
$2)$ If $|z|<1$ write $z=r(cosalpha+isinalpha)$, with $r<1$ and $alpha in [0,2pi]$ and then $g(z)=z^5=r^5(cos5alpha+isin5alpha))$ with, $r^5<1$ and $5alpha in [0,5pi]$.
answered Dec 6 '16 at 14:07
ArnaldoArnaldo
18.2k42246
$endgroup$
I also agree with your instructor.
$1)$ The graph of $f$ is a parabola with minimal point at $f(2)=-5$ and up concavity. See that $f(1)=-4$:
$A=(1,0)$ give us the point $B=(1,f(1)=-4)$ and then is easy to see that $f((1,infty))=[-5,infty[$. For $f^{-1}((1,infty))$ we have to see what are the values of $x$ such that $f(x) in (1,infty)$. The point $F$ represent the initial point. It is the point such that
$f(x)=1 Rightarrow x^2-4x-1=1 Rightarrow x=2+sqrt{6}$ and $x=2-sqrt{6}$
That means if we take $x<2-sqrt{6}$ or $x>2+sqrt{6}$ we will get $f(x) in (1,infty)$.
$2)$ If $|z|<1$ write $z=r(cosalpha+isinalpha)$, with $r<1$ and $alpha in [0,2pi]$ and then $g(z)=z^5=r^5(cos5alpha+isin5alpha))$ with, $r^5<1$ and $5alpha in [0,5pi]$.
answered Dec 6 '16 at 14:07
ArnaldoArnaldo
18.2k42246
edited Dec 6 '16 at 14:13
answered Dec 6 '16 at 14:07
ArnaldoArnaldo
18.2k42246
answered Dec 6 '16 at 14:07
ArnaldoArnaldo
18.2k42246
answered Dec 6 '16 at 14:07
ArnaldoArnaldo
18.2k42246
18.2k42246
add a comment |
add a comment |
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$begingroup$
Does $f^{-1}$ mean inverse of $f$?
$endgroup$
– Arnaldo
Dec 6 '16 at 12:54
$begingroup$
No, its the preimage .
$endgroup$
– Raducu Mihai
Dec 6 '16 at 13:07