RESTRICTION D'UNE APPLICATION
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HallsofIvy
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It's been a while since I read French but I'll give it a shot:
"Suppose I is the interval [0, 1) and J is the interval \(\displaystyle (1,\infty)\).
Let h be the function from \(\displaystyle I\cup J\) to R such that
\(\displaystyle h(x)= 1+ \frac{2}{\sqrt{x}- 1}\).
1. Is h injective? Surjective?
2. Let f be the restriction of h to J and g be the restriction of h to I.
(a) Show that, for all x in I, except x= 0, g(x)= -f(1/x).
(b) Show that f is a bijection from J to J.
(c) Show that g is a bijection from I to \(\displaystyle [-\infty, -1]\) and that its inverse is
\(\displaystyle g(x)= \left( \begin{array}{cc} g^{-1}(x)= 0 if x= -1 \\ g^{-1}(x)= \frac{1}{f^{-1}(-x)} if x<-1\end{array}\right)\)
"Suppose I is the interval [0, 1) and J is the interval \(\displaystyle (1,\infty)\).
Let h be the function from \(\displaystyle I\cup J\) to R such that
\(\displaystyle h(x)= 1+ \frac{2}{\sqrt{x}- 1}\).
1. Is h injective? Surjective?
2. Let f be the restriction of h to J and g be the restriction of h to I.
(a) Show that, for all x in I, except x= 0, g(x)= -f(1/x).
(b) Show that f is a bijection from J to J.
(c) Show that g is a bijection from I to \(\displaystyle [-\infty, -1]\) and that its inverse is
\(\displaystyle g(x)= \left( \begin{array}{cc} g^{-1}(x)= 0 if x= -1 \\ g^{-1}(x)= \frac{1}{f^{-1}(-x)} if x<-1\end{array}\right)\)
HallsofIvy
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.
If \(\displaystyle 1+ \frac{2}{\sqrt{x}- 1}= 1+ \frac{2}{\sqrt{y}- 1}\) then
\(\displaystyle \frac{2}{\sqrt{x}- 1}= \frac{2}{\sqrt{y}- 1}\)
"Injective", also called "one-to-one" means that if h(x)= h(y) then x= y
\(\displaystyle \sqrt{x}- 1= \sqrt{y}- 1\)
\(\displaystyle \sqrt{x}= \sqrt{y}\)
x= y. Yes, h is injective.
"Surjective", also called "onto", means that for every y there exists x such that \(\displaystyle y= 1+ \frac{2}{\sqrt{x}- 1}\). Essentially solve for x. \(\displaystyle y- 1= \frac{2}{\sqrt{x}- 1}\)
\(\displaystyle \sqrt{x}- 1= \frac{2}{y- 1}\)
\(\displaystyle \sqrt{x}= \frac{2}{y-1}+ 1\)
\(\displaystyle x= (\frac{2}{y-1}+ 1)^2\)
(Since the fraction, 2/anything, is never 0, y is never 1.)
Yes, h is surjective.
is y=(√x +1)/(√x - 1) one-to-one function? - Wolfram|Alpha
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