Cfrgtkky

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.

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 $z^5=re^{it}$?

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 $5alpha in [0,5pi]$.