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.

  • $\begingroup$ 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

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 $0\leq r<1$ and $t\in [0,2\pi)$. Can you find $z\in D$ such that $z^5=re^{it}$?

  • $\begingroup$ 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 $2\in(1,\infty)$ but $-5\not \in [-4,+\infty)$. 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

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 $f(1)=-4$:

enter image description here $A=(1,0)$ 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 $z=r(\cos\alpha+i\sin\alpha)$, with $r<1$ and $\alpha \in [0,2\pi]$ and then $g(z)=z^5=r^5(\cos5\alpha+i\sin5\alpha))$ with, $r^5<1$ and $5\alpha \in [0,5\pi]$.


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