Question

Help me, please. I confused
f:R\0 --> R ,
f(x) = 1/x^2
E =[1,2]
find f(E) and f^- (E)

Answer 1

if
\[\large 1\leq x\leq2 \]
then
\[\large 1\leq x^2\leq4 \]
\[\large 1\geq\frac{1}{x^2}\geq\frac{1}{4} \]
so
\[\large f(E)=[1/4;1] \]

Answer 2

I see.

Answer 3

OK. it makes no sense to me, but i understand.
U have a typo:
\[\large g^-(E)=[-2;-1]\cup[1;2] \]

Answer 4

yes. so u would get
\[\large f^-(E)=[-1;-1/\sqrt{2}]\cup[1/\sqrt{2};1] \]

Answer 5

\[\large g^-(E)=[-\sqrt{2};-1]\cup[1;\sqrt{2}] \]
\[\large h^-(g^-(E))=h^-([-\sqrt{2};-1]\cup[1;\sqrt{2}])= \]
\[\large =h^-([-\sqrt{2};-1])\cup h^-([1;\sqrt{2}]) \]
\[\large =[-1;1/\sqrt{2}]\cup[1/\sqrt{2};1] \]

Answer 6

sorry. but i have to go.

Answer 7

let g(x) = 1/x , so if E = \(\{ 1\leq x\leq 2\}\), then g(E) = \(\{ 1/2\leq x\leq 1\}=[1/2,1]\)
\(g^-(x) = 1/x \) so \(g^-(E) = [1/2,1]\)
am I right?

Answer 8

I feel bad I stuck around here for awhile lol. But what you wrote seems right to me so far. I'm not sure how helder defined g(x) in their response? is it the same g(x) you used

Answer 9

Pleaaaaaaaaaaaase, check,

Answer 10

:) @kirbykirby I got mad myself when I can't solve it by myself.

Answer 11

. The way I did it was the following, and gave the same result as @helder_edwin

\(1\le x \le 2\),


Let \(\dfrac{1}{x^2}=f(x) \implies\pm \sqrt{\dfrac{1}{x}}=f^{-1}(x)\)

\(\implies \dfrac{1}{\sqrt{x}}\) OR \( -\dfrac{1}{\sqrt{x}}\) so,

consider the positive branch first ... transform your interval into the given format.

\(1\le x \le 2 \implies 1\ge \dfrac{1}{x}\ge \dfrac{1}{2} \\
~ \\ \implies 1 \ge \dfrac{1}{\sqrt{x}} \ge \dfrac{1}{\sqrt{2}} \\
\implies \dfrac{1}{\sqrt{2}}\ \le \dfrac{1}{\sqrt{x}} \le 1\)

Now the negative branch: It's the same process, but at the last line:
\(-\dfrac{1}{\sqrt{2}}\ \ge -\dfrac{1}{\sqrt{x}} -\ge 1\)
\( \implies -1 \le -\dfrac{1}{\sqrt{x}} \le -\dfrac{1}{\sqrt{2}}\)

So overall: \(\left[-1,-\dfrac{1}{\sqrt{2}}\right] \cup \left[\dfrac{1}{\sqrt{2}},1 \right]\)

Answer 12

@Loser66

Answer 13

(your attachment wouldn't load for some reason o.O)

Answer 14

And it is the same answer with me, yeahhhhhhhhhhhhhh. Thank you

Answer 15

ah good :) awesome